Quadratic Formula Proof

Suppose we have a quadratic, \(ax^2+bx+c=0\) . We can solve for \(x\) by completing the square and manipulating the equation as follows:

\[ax^2+bx+c=0\]

\[a\left(x^2+\frac{b}{a}x\right)+c=0\]

\[\left(x^2+\frac{b}{a}x\right)+\frac{c}{a}=0\]

\[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0\]

\[\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\]

\[\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\]

\[x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\]

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

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

\[\therefore x=\frac{                 -b \pm \sqrt{b^2-4ac}               }{2a}\]

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