Theorem

The quadratic equation of the form $a x^2 + b x + c = 0$ has solutions:

$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$

Real Coefficients

Let $a, b, c \in \R$.

The quadratic equation $a x^2 + b x + c = 0$ has:

Two real solutions if $b^2 - 4 a c > 0$
One real solution if $b^2 - 4 a c = 0$
Two complex solutions if $b^2 - 4 a c < 0$, and those two solutions are complex conjugates.

Proof

Let $a x^2 + b x + c = 0$. Then:

 $\displaystyle 4 a^2 x^2 + 4 a b x + 4 a c$ $=$ $\displaystyle 0$ multiplying through by $4 a$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {2 a x + b}^2 - b^2 + 4 a c$ $=$ $\displaystyle 0$ Completing the Square $\displaystyle \leadsto \ \$ $\displaystyle \paren {2 a x + b}^2$ $=$ $\displaystyle b^2 - 4 a c$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \frac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$

$\blacksquare$

Also known as

This result is often referred to as the quadratic formula.