(5th millennium BC - 4th millennium BC - 3rd millennium BC)
Significant persons:
Inventions, Discoveries, Introductions:
... 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> ...