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**SuperCalculus** So, I was given the task of proving the quadratic formula for homework... I think I've pretty much got it, but I'd like to be sure, so if someone can look over my working it'd be much appreciated.

$\displaystyle ax^2 + bx + c = 0$

Completing the square gives;

$\displaystyle a(x + \frac{b^2}{2a})^2 - \frac{b^2}{4a} + c = 0$

$\displaystyle a(x + \frac{b^2}{2a})^2 = \frac{b^2}{4a} - c$

$\displaystyle a(x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a}$

$\displaystyle (x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$

$\displaystyle (x + \frac{b^2}{2a}) = \frac{\sqrt {b^2 - 4ac}}{\sqrt{4}\sqrt{a^2}} = \frac{\sqrt {b^2 - 4ac}}{2a} $

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