(21st century - 22nd century - 23rd century - other centuries)
Decades and years
... 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 ...