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