quadratic

Convert vertex form to standard form: $2(x-4)^2 + 8 = 2(x^2 - 8x + 16) + 8 = 2x^2- 16x + 40$

This means that $a = 2$, $b = -16$, and $c = 40$. So, the quadratic we need to convert to vertex form is $6x^2 - 32x + 40$.

Note that in vertex form (which we need to convert it to), the vertex occurs at $(h,k)$.

This means we need to find the x-coordinate of the vertex, which occurs at ${-b \over 2a}={32 \over 12} = \color{brown}\boxed{8 \over 3}$.