User:DukeEgr93/Quadratic Equation

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:

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