if we were to use the quadratic formula, we would get that the roots are
[tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
or that the 2 seperate roots are
[tex]\frac{-b+\sqrt{b^2-4ac}}{2a}[/tex] and [tex]\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]
if we sum these, then the [tex]\sqrt{b^2-4ac}[/tex] bits will cancel and we wil be left with
[tex]\frac{-b-b}{2a}[/tex] or [tex]\frac{-2b}{2a}[/tex] or [tex]\frac{-b}{a}[/tex]
the sum of the solutions is [tex]\frac{-b}{a}[/tex]
