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# MathML: Deriving the Quadratic Formula

$\begin{array}{l}a{x}^{2}+bx+c=0\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}a{x}^{2}+bx\phantom{\rule[-1ex]{2.5ex}{0.5ex}}=-c\phantom{\rule[-1ex]{2.5ex}{0.5ex}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}{x}^{2}+\frac{b}{a}⁤x\phantom{\rule[-1ex]{2.5ex}{0.5ex}}=\frac{-c}{a}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{Divide out leading coefficient.}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}{x}^{2}+\frac{b}{a}⁤x+{\left(\frac{b}{2a}\right)}^{2}=\frac{-c\left(4a\right)}{a\left(4a\right)}+\frac{{b}^{2}}{4{a}^{2}}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{Complete the square.}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}\left(x+\frac{b}{2a}\right)\left(x+\frac{b}{2a}\right)=\frac{{b}^{2}-4ac}{4{a}^{2}}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{Discriminant revealed.}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}x+\frac{b}{2a}=\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}x=\frac{-b}{2a}±\left\{C\right\}\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{There's the vertex formula.}}\\ \phantom{\rule[-1ex]{2.5ex}{0.5ex}}x=\frac{-b±\left\{C\right\}\sqrt{{b}^{2}-4ac}}{2a}\phantom{\rule[-1ex]{2ex}{0.5ex}}{\text{}}\end{array}$