December 2002
November 2002
October 2002
September 2002
August 2002
July 2002
June 2002
May 2002
April 2002
March 2002
January 2002
... 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 formula and its proof ...