User:DukeEgr93/Quadratic Equation

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How do you get from:


ax^2+bx+c=0\,\!

to


x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\,\!

? Other than practice, practice, practice - it's completing the square:


\begin{align}
ax^2+bx+c&=0\\
x^2+\frac{b}{a}x+\frac{c}{a}&=0\\
x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}&=0\\
x^2+\frac{b}{a}x+\frac{b^2}{4a}&=\frac{b^2}{4a^2}-\frac{c}{a}\\
\left(x+\frac{b}{2a}\right)^2&=\left(\frac{b^2-4ac}{4a^2}\right)\\
x+\frac{b}{2a}&=\pm\sqrt{\left(\frac{b^2-4ac}{4a^2}\right)}\\
x&=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{b^2-4ac}\\
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\end{align}