Solution to the Quadratic Equation

Theorem 8:  If $ax^2 + bx + c=0$, then:

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

Proof:

(1)  $x^2 + \dfrac{bx}{a} + \dfrac{c}{a} = 0$

(2)  $x^2 + \dfrac{bx}{a} = -\dfrac{c}{a}$

(3)  $x^2 + \dfrac{bx}{a} + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(4)  $\left(x + \dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(5)  $x + \dfrac{b}{2a} = \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2}$

(6) $x = -\dfrac{b}{2a} \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2} = \dfrac{-b \pm  \sqrt{b^2 - 4ac}}{2a}$