\displaystyle \text{Question 1: The distance by road between two towns A and B is }216\text{ km, and by rail, it is }208\text{ km.}
\displaystyle \text{A car travels at a speed of }x\text{ km/hr and the train travels at a speed which is }16\text{ km/hr faster than the car. Calculate:}
\displaystyle \text{(i) The time taken by the car to reach town B from A, in terms of }x.
\displaystyle \text{(ii) The time taken by the train to reach town B from A, in terms of }x.
\displaystyle \text{(iii) If the train takes }2\text{ hours less than the car to reach town B, obtain an equation in }x\text{ and solve it.}
\displaystyle \text{Hence, find the speed of the train.}\hfill \text{[ICSE 1998]}
\displaystyle \text{Answer:}
\displaystyle \text{Speed of car }=x\text{ km/hr}
\displaystyle \text{Speed of train }=(x+16)\text{ km/hr}
\displaystyle \text{(i) Time taken by the car }=\frac{216}{x}\text{ hrs}
\displaystyle \text{(ii) Time taken by the train }=\frac{208}{x+16}\text{ hrs}
\displaystyle \text{Given, train takes }2\text{ hours less than the car.}
\displaystyle \frac{216}{x}-\frac{208}{x+16}=2
\displaystyle \frac{216(x+16)-208x}{x(x+16)}=2
\displaystyle 216x+3456-208x=2x(x+16)
\displaystyle 8x+3456=2x^2+32x
\displaystyle 2x^2+24x-3456=0
\displaystyle x^2+12x-1728=0
\displaystyle (x-36)(x+48)=0
\displaystyle x=36\text{ or }x=-48
\displaystyle \text{Since speed cannot be negative, }x=36
\displaystyle \therefore \text{Speed of train }=x+16=36+16=52\text{ km/hr}
\\

\displaystyle \text{Question 2: A trader buys }x\text{ articles for a total cost of Rs. }600.
\displaystyle \text{Write down the cost of one article in terms of }x.\text{ If the cost per article were Rs. }5\text{ more,}
\displaystyle \text{the number of articles that can be bought for Rs. }600\text{ would be }4\text{ less. Write down the equation in }x
\displaystyle \text{for the above situation and solve it for }x.\hfill \text{[ICSE 1999]}
\displaystyle \text{Answer:}
\displaystyle \text{Cost of one article }=\text{Rs. }\frac{600}{x}
\displaystyle \text{New cost of one article }=\text{Rs. }\left(\frac{600}{x}+5\right)
\displaystyle \text{New number of articles }=x-4
\displaystyle \frac{600}{\left(\frac{600}{x}+5\right)}=x-4
\displaystyle 600=\left(x-4\right)\left(\frac{600}{x}+5\right)
\displaystyle 600x=(x-4)(600+5x)
\displaystyle 600x=600x+5x^2-2400-20x
\displaystyle 5x^2-20x-2400=0
\displaystyle x^2-4x-480=0
\displaystyle (x-24)(x+20)=0
\displaystyle x=24\text{ or }x=-20
\displaystyle \text{Since the number of articles cannot be negative, }x=24
\displaystyle \therefore \text{Number of articles bought }=24
\\

\displaystyle \text{Question 3: A hotel bill for a number of people for an overnight stay is Rs. }4800.
\displaystyle \text{If there were }4\text{ people more, the bill each person had to pay would have reduced by Rs. }200.
\displaystyle \text{Find the number of people staying overnight.}\hfill \text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of people staying overnight }=x
\displaystyle \text{Bill paid by each person }=\text{Rs. }\frac{4800}{x}
\displaystyle \text{If there were }4\text{ people more, number of people }=x+4
\displaystyle \text{New bill paid by each person }=\text{Rs. }\frac{4800}{x+4}
\displaystyle \frac{4800}{x}-\frac{4800}{x+4}=200
\displaystyle \frac{4800(x+4)-4800x}{x(x+4)}=200
\displaystyle 4800x+19200-4800x=200x(x+4)
\displaystyle 19200=200x^2+800x
\displaystyle x^2+4x-96=0
\displaystyle (x-8)(x+12)=0
\displaystyle x=8\text{ or }x=-12
\displaystyle \text{Since the number of people cannot be negative, }x=8
\displaystyle \therefore \text{Number of people staying overnight }=8
\\

\displaystyle \text{Question 4: An airplane travelled a distance of }400\text{ km at an average speed of }x\text{ km/hr.}
\displaystyle \text{On the return journey, the speed was increased by }40\text{ km/hr. Write down expressions for the time taken}
\displaystyle \text{for the onward journey and the return journey. If the airplane takes }2\text{ hours less in returning,}
\displaystyle \text{calculate the speed of the airplane.}\hfill \text{[ICSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{Time taken for onward journey }=\frac{400}{x}\text{ hrs}
\displaystyle \text{Time taken for return journey }=\frac{400}{x+40}\text{ hrs}
\displaystyle \frac{400}{x}-\frac{400}{x+40}=2
\displaystyle \frac{400(x+40)-400x}{x(x+40)}=2
\displaystyle 400x+16000-400x=2x(x+40)
\displaystyle 16000=2x^2+80x
\displaystyle x^2+40x-8000=0
\displaystyle x=\frac{-40\pm\sqrt{40^2-4(1)(-8000)}}{2}
\displaystyle x=\frac{-40\pm\sqrt{33600}}{2}
\displaystyle x=\frac{-40\pm183.30}{2}
\displaystyle x=71.65\text{ or }x=-111.65
\displaystyle \text{Since speed cannot be negative, }x=71.65
\displaystyle \therefore \text{Speed of the airplane }=71.65\text{ km/hr}
\\

\displaystyle \text{Question 5: In an auditorium, seats were arranged in rows and columns. The number of rows was equal}
\displaystyle \text{to the number of seats in each row. When the number of rows was doubled and the number of seats}
\displaystyle \text{in each row was reduced by }10,\text{ the total number of seats increased by }300.\text{ Find the number of rows}
\displaystyle \text{in the original arrangement and the number of seats in the auditorium after re-arrangement.}\hfill \text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of rows in the original arrangement }=x
\displaystyle \text{Then, number of seats in each row }=x
\displaystyle \text{Original number of seats }=x\times x=x^2
\displaystyle \text{New number of rows }=2x
\displaystyle \text{New number of seats in each row }=x-10
\displaystyle \text{New number of seats }=2x(x-10)
\displaystyle 2x(x-10)-x^2=300
\displaystyle 2x^2-20x-x^2=300
\displaystyle x^2-20x-300=0
\displaystyle (x-30)(x+10)=0
\displaystyle x=30\text{ or }x=-10
\displaystyle \text{Since the number of rows cannot be negative, }x=30
\displaystyle \therefore \text{Number of rows in the original arrangement }=30
\displaystyle \text{Number of seats after re-arrangement }=2x(x-10)=2(30)(20)=1200
\\

\displaystyle \text{Question 6: Rs. }480\text{ is divided equally among }x\text{ children. If the number of children were }20\text{ more,}
\displaystyle \text{then each would have got Rs. }12\text{ less. Find }x.\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of children }=x
\displaystyle \text{Amount received by each child }=\text{Rs. }\frac{480}{x}
\displaystyle \text{If the number of children were }20\text{ more, number of children }=x+20
\displaystyle \text{New amount received by each child }=\text{Rs. }\frac{480}{x+20}
\displaystyle \frac{480}{x}-\frac{480}{x+20}=12
\displaystyle \frac{480(x+20)-480x}{x(x+20)}=12
\displaystyle 480x+9600-480x=12x(x+20)
\displaystyle 9600=12x^2+240x
\displaystyle x^2+20x-800=0
\displaystyle (x-20)(x+40)=0
\displaystyle x=20\text{ or }x=-40
\displaystyle \text{Since the number of children cannot be negative, }x=20
\displaystyle \therefore \text{Number of children }=20
\\

\displaystyle \text{Question 7: A car covers a distance of }400\text{ km at a certain speed. Had the speed been }12\text{ km/hr more,}
\displaystyle \text{the time taken for the journey would have been }1\text{ hour }40\text{ minutes less. Find the original speed of the car.}\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original speed of the car }=x\text{ km/hr}
\displaystyle \text{Original time taken }=\frac{400}{x}\text{ hrs}
\displaystyle \text{Increased speed }=(x+12)\text{ km/hr}
\displaystyle \text{New time taken }=\frac{400}{x+12}\text{ hrs}
\displaystyle 1\text{ hour }40\text{ minutes}=1+\frac{40}{60}=\frac{5}{3}\text{ hrs}
\displaystyle \frac{400}{x}-\frac{400}{x+12}=\frac{5}{3}
\displaystyle \frac{400(x+12)-400x}{x(x+12)}=\frac{5}{3}
\displaystyle 3[400x+4800-400x]=5x(x+12)
\displaystyle 14400=5x^2+60x
\displaystyle 5x^2+60x-14400=0
\displaystyle x^2+12x-2880=0
\displaystyle (x-48)(x+60)=0
\displaystyle x=48\text{ or }x=-60
\displaystyle \text{Since speed cannot be negative, }x=48
\displaystyle \therefore \text{Original speed of the car }=48\text{ km/hr}
\\

\displaystyle \text{Question 8: A positive number is divided into two parts such that the sum of the squares of the two parts is }20.
\displaystyle \text{The square of the larger part is }8\text{ times the smaller part. Taking }x\text{ as the smaller part, find the number.}\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the smaller part }=x
\displaystyle \text{Let the larger part }=y
\displaystyle x^2+y^2=20
\displaystyle y^2=8x
\displaystyle x^2+8x=20
\displaystyle x^2+8x-20=0
\displaystyle (x+10)(x-2)=0
\displaystyle x=-10\text{ or }x=2
\displaystyle \text{Since the number is positive, }x=2
\displaystyle y^2=8x=8(2)=16
\displaystyle y=4
\displaystyle \therefore \text{Required number }=x+y=2+4=6
\\

\displaystyle \text{Question 9: By increasing the speed of a car by }10\text{ km/hr, the time of the journey for a distance of }72\text{ km}
\displaystyle \text{is reduced by }36\text{ minutes. Find the original speed of the car.}\hfill \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original speed of the car }=x\text{ km/hr}
\displaystyle \text{Original time taken }=\frac{72}{x}\text{ hrs}
\displaystyle \text{New speed of the car }=(x+10)\text{ km/hr}
\displaystyle \text{New time taken }=\frac{72}{x+10}\text{ hrs}
\displaystyle 36\text{ minutes}=\frac{36}{60}=\frac{3}{5}\text{ hrs}
\displaystyle \frac{72}{x}-\frac{72}{x+10}=\frac{3}{5}
\displaystyle \frac{72(x+10)-72x}{x(x+10)}=\frac{3}{5}
\displaystyle 5(720)=3x(x+10)
\displaystyle 3600=3x^2+30x
\displaystyle x^2+10x-1200=0
\displaystyle (x-30)(x+40)=0
\displaystyle x=30\text{ or }x=-40
\displaystyle \text{Since speed cannot be negative, }x=30
\displaystyle \therefore \text{Original speed of the car }=30\text{ km/hr}
\\

\displaystyle \text{Question 10: A two-digit number is such that the product of the digits is }6.\text{ When }9\text{ is added}
\displaystyle \text{to the number, the digits interchange their places. Find the number.}\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the ten's digit be }x\text{ and the unit's digit be }y
\displaystyle \text{Required number }=10x+y
\displaystyle \text{Number obtained by interchanging the digits }=10y+x
\displaystyle xy=6
\displaystyle 10x+y+9=10y+x
\displaystyle 10x-x+y-10y=-9
\displaystyle 9x-9y=-9
\displaystyle x-y=-1
\displaystyle y=x+1
\displaystyle x(x+1)=6
\displaystyle x^2+x-6=0
\displaystyle (x+3)(x-2)=0
\displaystyle x=-3\text{ or }x=2
\displaystyle \text{Since }x\text{ is a digit, }x=2
\displaystyle y=x+1=2+1=3
\displaystyle \therefore \text{Required number }=10x+y=10(2)+3=23
\\

\displaystyle \text{Question 11: Five years ago, a woman's age was the square of her son's age. }10\text{ years hence,}
\displaystyle \text{her age will be twice that of her son's age. Find: (i) the age of her son five years ago,}
\displaystyle \text{(ii) the present age of the woman.}\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the age of the son }5\text{ years ago }=x\text{ years}
\displaystyle \text{Woman's age }5\text{ years ago }=x^2\text{ years}
\displaystyle \text{Present age of the son }=(x+5)\text{ years}
\displaystyle \text{Present age of the woman }=(x^2+5)\text{ years}
\displaystyle \text{Age of the son }10\text{ years hence }=(x+15)\text{ years}
\displaystyle \text{Age of the woman }10\text{ years hence }=(x^2+15)\text{ years}
\displaystyle x^2+15=2(x+15)
\displaystyle x^2+15=2x+30
\displaystyle x^2-2x-15=0
\displaystyle (x-5)(x+3)=0
\displaystyle x=5\text{ or }x=-3
\displaystyle \text{Since age cannot be negative, }x=5
\displaystyle \therefore \text{(i) Age of her son }5\text{ years ago }=5\text{ years}
\displaystyle \text{(ii) Present age of the woman }=x^2+5=25+5=30\text{ years}
\\

\displaystyle \text{Question 12: A shopkeeper buys a certain number of books for Rs. }960.\text{ If the cost per book was Rs. }8\text{ less,}
\displaystyle \text{the number of books that he could have bought for Rs. }960\text{ would be }4\text{ more. Taking the original cost}
\displaystyle \text{of each book to be Rs. }x,\text{ write an equation in }x\text{ and solve it.}\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original cost of each book }=\text{Rs. }x
\displaystyle \text{Number of books bought }=\frac{960}{x}
\displaystyle \text{If the cost per book was Rs. }8\text{ less, new cost }=\text{Rs. }(x-8)
\displaystyle \text{New number of books bought }=\frac{960}{x-8}
\displaystyle \frac{960}{x-8}-\frac{960}{x}=4
\displaystyle \frac{960x-960(x-8)}{x(x-8)}=4
\displaystyle \frac{960x-960x+7680}{x(x-8)}=4
\displaystyle \frac{7680}{x(x-8)}=4
\displaystyle 7680=4x(x-8)
\displaystyle 1920=x^2-8x
\displaystyle x^2-8x-1920=0
\displaystyle (x-48)(x+40)=0
\displaystyle x=48\text{ or }x=-40
\displaystyle \text{Since cost cannot be negative, }x=48
\displaystyle \therefore \text{Original cost of each book }=\text{Rs. }48
\\

\displaystyle \text{Question 13: Some students planned a picnic. The budget for the food was Rs. }480.\text{ As eight of the students}
\displaystyle \text{failed to join the party, the cost of the food for each member increased by Rs. }10.\text{ Find how many students}
\displaystyle \text{went for the picnic.}\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students who planned the picnic }=x
\displaystyle \text{Planned food budget }=\text{Rs. }480
\displaystyle \text{Planned budget per student }=\text{Rs. }\frac{480}{x}
\displaystyle \text{Number of students who actually went for the picnic }=x-8
\displaystyle \text{Actual food budget per student }=\text{Rs. }\frac{480}{x-8}
\displaystyle \frac{480}{x-8}-\frac{480}{x}=10
\displaystyle \frac{480x-480(x-8)}{x(x-8)}=10
\displaystyle \frac{480x-480x+3840}{x(x-8)}=10
\displaystyle \frac{3840}{x(x-8)}=10
\displaystyle 3840=10x(x-8)
\displaystyle 384=x^2-8x
\displaystyle x^2-8x-384=0
\displaystyle (x-24)(x+16)=0
\displaystyle x=24\text{ or }x=-16
\displaystyle \text{Since the number of students cannot be negative, }x=24
\displaystyle \therefore \text{Number of students who actually went for the picnic }=24-8=16
\\

Solve each of the following equations for \displaystyle x  

\displaystyle \text{Question 14: Solve }x^2-3x-9=0.\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }x^2-3x-9=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=-3,\ c=-9
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-9)}}{2(1)}
\displaystyle x=\frac{3\pm\sqrt{9+36}}{2}
\displaystyle x=\frac{3\pm\sqrt{45}}{2}
\displaystyle x=\frac{3\pm6.708}{2}
\displaystyle x=4.85\text{ or }x=-1.85
\\

\displaystyle \text{Question 15: Solve }x^2-5x-10=0.\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }x^2-5x-10=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=-5,\ c=-10
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-10)}}{2(1)}
\displaystyle x=\frac{5\pm\sqrt{25+40}}{2}
\displaystyle x=\frac{5\pm\sqrt{65}}{2}
\displaystyle x=\frac{5\pm8.062}{2}
\displaystyle x=6.53\text{ or }x=-1.53
\\

\displaystyle \text{Question 16: Solve }2x-\frac{1}{x}=7.\hfill \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle 2x-\frac{1}{x}=7
\displaystyle 2x^2-1=7x
\displaystyle 2x^2-7x-1=0
\displaystyle \text{Comparing }2x^2-7x-1=0\text{ with }ax^2+bx+c=0,
\displaystyle a=2,\ b=-7,\ c=-1
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-1)}}{2(2)}
\displaystyle x=\frac{7\pm\sqrt{49+8}}{4}
\displaystyle x=\frac{7\pm\sqrt{57}}{4}
\displaystyle x=\frac{7\pm7.55}{4}
\displaystyle x=3.64\text{ or }x=-0.14
\\

\displaystyle \text{Question 17: Solve }5x^2-3x-4=0.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }5x^2-3x-4=0\text{ with }ax^2+bx+c=0,
\displaystyle a=5,\ b=-3,\ c=-4
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(5)(-4)}}{2(5)}
\displaystyle x=\frac{3\pm\sqrt{9+80}}{10}
\displaystyle x=\frac{3\pm\sqrt{89}}{10}
\displaystyle x=\frac{3\pm9.434}{10}
\displaystyle x=1.243\text{ or }x=-0.643
\displaystyle \therefore \text{Correct to three significant figures, }x=1.24\text{ or }x=-0.643
\\

\displaystyle \text{Question 18: Solve }(x-1)^2-3x+4=0.\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle (x-1)^2-3x+4=0
\displaystyle x^2-2x+1-3x+4=0
\displaystyle x^2-5x+5=0
\displaystyle \text{Comparing }x^2-5x+5=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=-5,\ c=5
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(5)}}{2(1)}
\displaystyle x=\frac{5\pm\sqrt{25-20}}{2}
\displaystyle x=\frac{5\pm\sqrt{5}}{2}
\displaystyle x=\frac{5\pm2.236}{2}
\displaystyle x=3.618\text{ or }x=1.382
\displaystyle \therefore \text{Correct to two significant figures, }x=3.6\text{ or }x=1.4
\\

\displaystyle \text{Question 19: Solve }x^2-5x-10=0.\hfill \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }x^2-5x-10=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=-5,\ c=-10
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-10)}}{2(1)}
\displaystyle x=\frac{5\pm\sqrt{25+40}}{2}
\displaystyle x=\frac{5\pm\sqrt{65}}{2}
\displaystyle x=\frac{5\pm8.062}{2}
\displaystyle x=6.53\text{ or }x=-1.53
\\

\displaystyle \text{Question 20: Solve }3x^2-x-7=0.\hfill \text{[ICSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }3x^2-x-7=0\text{ with }ax^2+bx+c=0,
\displaystyle a=3,\ b=-1,\ c=-7
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-7)}}{2(3)}
\displaystyle x=\frac{1\pm\sqrt{1+84}}{6}
\displaystyle x=\frac{1\pm\sqrt{85}}{6}
\displaystyle x=\frac{1\pm9.220}{6}
\displaystyle x=1.703\text{ or }x=-1.370
\\

\displaystyle \text{Question 21: Without solving the following quadratic equation, find the value of }p\text{ for which}
\displaystyle \text{the roots are equal: }px^2-4x+3=0.\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }px^2-4x+3=0\text{ with }ax^2+bx+c=0,
\displaystyle a=p,\ b=-4,\ c=3
\displaystyle \text{For equal roots, }b^2-4ac=0
\displaystyle (-4)^2-4(p)(3)=0
\displaystyle 16-12p=0
\displaystyle 12p=16
\displaystyle p=\frac{16}{12}=\frac{4}{3}
\\

\displaystyle \text{Question 22: Find the value of }m\text{ for which the roots of }x^2+2(m-1)x+(m+5)=0\text{ are equal.}\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }x^2+2(m-1)x+(m+5)=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=2(m-1),\ c=m+5
\displaystyle \text{For equal roots, }b^2-4ac=0
\displaystyle [2(m-1)]^2-4(1)(m+5)=0
\displaystyle 4(m-1)^2-4(m+5)=0
\displaystyle (m-1)^2-(m+5)=0
\displaystyle m^2-2m+1-m-5=0
\displaystyle m^2-3m-4=0
\displaystyle (m-4)(m+1)=0
\displaystyle m=4\text{ or }m=-1
\\

\displaystyle \text{Question 23: Solve the following equation: }x-\frac{18}{x}=6.\text{ Give your answer correct to two significant figures.}\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle x-\frac{18}{x}=6
\displaystyle x^2-18=6x
\displaystyle x^2-6x-18=0
\displaystyle \text{Comparing }x^2-6x-18=0\text{ with }ax^2+bx+c=0,
\displaystyle a=1,\ b=-6,\ c=-18
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{6\pm\sqrt{(-6)^2-4(1)(-18)}}{2(1)}
\displaystyle x=\frac{6\pm\sqrt{36+72}}{2}
\displaystyle x=\frac{6\pm\sqrt{108}}{2}
\displaystyle x=\frac{6\pm6\sqrt{3}}{2}
\displaystyle x=3\pm3\sqrt{3}
\displaystyle x=8.196\text{ or }x=-2.196
\displaystyle \therefore \text{Correct to two significant figures, }x=8.2\text{ or }x=-2.2
\\

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