When \(a \ne 0\) , there are two solutions to \(ax^2 + bx + c = 0\) and they are

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

Discriminant: \(D = b^2-4ac\)

If \(\alpha\) and \(\beta\) are the roots of quadratic equation (\(ax^2 + bx + c = 0\)), then:

- \(\alpha + \beta = -{b \over a}\)
- \(\alpha \beta = {c \over a}\)
- \((x - \alpha)(x - \beta) = 0\)