This article (or an earlier version of it) contains material from FOLDOC, used with permission. Update as needed.
... b/(2a) from both sides, we get <math>x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math> Generalizations The ...