Derive quadratic equation solution

\[ax^2+bx+c=0\,\!\]
\[ax^2+bx=-c\,\!\]

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Convert \(ax^2+bx\) to the form \(u^2x^2+2uvx+v^2\). Ignore the last term (for now) and let \(v=b\). If we start with \(u=a\) we can multiply both sides by \(a\).

\[a^2x^2+abx=-ac\,\!\]Not quite the form needed. Still missing a \(2\) in the second term. We need a number such that \(2d=d^2\) which is clearly \(d=2\). This gives \(u=2a\) which can be achieved by multiplying both sides by \(4\)

\[4a^2x^2+4abx=-4ac\,\!\]Now add \(b^2\) to both sides:

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