I didn't know you were an Aussie. Add yourself to Wikipedia:Wikipedians/Australia. Welcome to Wikipedia. :) -- Tim Starling 07:01 30 Jun 2003 (UTC)
... 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 formul ...