Answer
$\frac{x_1+x_2}{2}=-\frac{b}{2a}$
Work Step by Step
$x=\frac{-b±\sqrt {b^2-4ac}}{2a}$
$x_1=-\frac{b}{2a}+\frac{\sqrt {b^2-4ac}}{2a}$
$x_2=-\frac{b}{2a}-\frac{\sqrt {b^2-4ac}}{2a}$
The average of the zeros:
$\frac{x_1+x_2}{2}=\frac{-\frac{b}{2a}+\frac{\sqrt {b^2-4ac}}{2a}-\frac{b}{2a}-\frac{\sqrt {b^2-4ac}}{2a}}{2}=\frac{-\frac{b}{a}}{2}=-\frac{b}{2a}$
Also the x-coordinate of the vertex is: $-\frac{b}{2a}$
