+1904 Can you expand it and show how you got to \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). I already know that that is the answer. I want to show why the quadradic formula works.
+37155 Here is a webpage with the derivation of the Quadratic Formula:
http://www.purplemath.com/modules/sqrquad2.htm
scroll down a little bit to see the derivation
+118687 Why didn't you say so LOL
\(\begin{array}{rcl}\\ax^2+bx+c&=&0\\ ax^2+bx&=&-c\\ x^2+\frac{b}{a}x&=&-\frac{c}{a}\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{c}{a}+(\frac{b}{2a})^2\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ (x+\frac{b}{2a})^2&=&\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=&\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \end{array}\)
