Solution to Quadratic Equation/Real Coefficients

Theorem
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
From Solution to Quadratic Equation:


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

If the discriminant $b^2 - 4 a c > 0$ then $\sqrt {b^2 - 4 a c}$ has two values and the result follows.

If the discriminant $b^2 - 4 a c = 0$ then $\sqrt {b^2 - 4 a c} = 0$ and $x = \dfrac {-b} {2 a}$.

If the discriminant $b^2 - 4 a c < 0$, then it can be written as:


 * $b^2 - 4 a c = \left({-1}\right) \left|{b^2 - 4 a c}\right|$

Thus:
 * $\sqrt {b^2 - 4 a c} = \pm i \sqrt {\left|{b^2 - 4 a c}\right|}$

and the two solutions are:


 * $x = \dfrac {-b} {2 a} + i \dfrac {\sqrt {\left|{b^2 - 4 a c}\right|}} {2 a}, x = \dfrac {-b} {2 a} - i \dfrac {\sqrt {\left|{b^2 - 4 a c}\right|}} {2 a}$

and once again the result follows.