It is freely available and defined in RFC 2144. http://www.faqs.org/rfcs/rfc2144
... 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 remain ...