Every polynomial equation $f(x) = 0$ has at least one root, real or imaginary ( complex).

Every polynomial equation $f(x) = 0$ of degree $n$ has exactly $n$ roots real or imaginary.

A quadratic equation cannot have more than two roots.

If $ax^2 + bx + c = 0$ is the quadratic equation we get, the roots

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \text{If } b^2 - 4ac = 0 \text{ then } \alpha = \beta \text{ and the roots are } \frac{-b}{2a}$

If $b^2 - 4ac > 0$ but is not a perfect square, then the roots are irrational and unequal.

If $b^2 - 4ac < 0$  then the roots are imaginary and are give by

$\displaystyle \alpha = \frac{-b + i\sqrt{4ac - b^2} }{2a} \text{ and } \beta = \frac{-b - i\sqrt{4ac - b^2} }{2a}$

Complex roots of an equation with real coefficient  always occur in pairs. However, this may not be true in case of the equations with complex coefficient. For example, $x^2 - 2ix - 1 = 0$ has both roots equal to $i$.

Surd root of an equation with rational coefficient  always occurs in pairs like $2 + \sqrt{3}$ and $2 - \sqrt{3}$. However, Surd roots of an equation with irrational coefficients may not occur in pairs. For example, $x^2 - 2 \sqrt{3} x + 3 = 0$ has both roots equal to $\sqrt{3}$.