I’m attempting to verify a limited relationship one of my high school students observed.

$(x_1-x_2)^2/4=(x_1+x_2)^2/4-x_1x_2 = (-b/2a)^2 -c/a = (b^2-4ac)/(4a^2)$

At the vertex, $x=b/2a$ and the y-coordinate is $(-b^2+4ac)/(4a)$

So the second is $-a$ times the first.

If you are familiar with differention, you may consider the function $f(x) =a(x-x_1)(x-x_2)$. Then the vertex of the parabola comes at the point where $f'(x)=0$ and that is $t=\frac{x_1+x_2}{2}$. So $f(t) =-a\frac{(x_1-x_2)^2}{4}$