Answer
The vertex form of the function is $y=(x-(-10))^{2}+(10).$ The vertex is $(-10,-10)$.
Work Step by Step
$ y=x^{2}+20x+90\qquad$ ...write in form of $x^{2}+bx=c$ (add $-90$ to each side).
$ y-90=x^{2}+20x\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{20}{2})^{2}=(10)^{2}=100\qquad$ ...complete the square by adding $100$ to each side of the expression
$ y-90+100=x^{2}+20x+100\qquad$ ... write $x^{2}+20x+100$ as a binomial squared.
$ y+10=(x+10)^{2}\qquad$ ...add $-10$ to each side of the expression
$ y=(x+10)^{2}-10\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$y=(x-(-10))^{2}+(10)$
The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph.
Here, $h=-10,\ k=-10$, so the vertex is $(-10,-10)$
