(2nd millennium BC - 1st millennium BC - 1st millennium AD - other millennia)
Significant persons:
Inventions, Discoveries, Introductions
Centuries and Decades
... 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> ...