Answer
The vertex form of the function is $y=5(x+1)^{2}+2.$ The vertex is $(-1,2)$.
Work Step by Step
$ y=5x^{2}+10x+7\qquad$ ...factor out $5$ from the first two terms.
$ y=5(x^{2}+2x)+7\qquad$ ...add $-7$ to each side
$ y-7=5(x^{2}+2x)\qquad$ ...divide the entire expression with $5$.
$\displaystyle \frac{y-7}{5}=x^{2}+2x\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{2}{2})^{2}=1^{2}=1\qquad$ ...complete the square by adding$ 1$ to each side of the expression
$\displaystyle \frac{y-7}{5}+1=x^{2}+2x+1\qquad$ ... write $x^{2}+2x+1$ as a binomial squared.
$\displaystyle \frac{y-7}{5}+1=(x+1)^{2}\qquad$ ...simplify.$\displaystyle \frac{y-7}{5}+1=\frac{y-7}{5}+\frac{5}{5}=\frac{y-7+5}{5}=\frac{y-2}{5}$
$\displaystyle \frac{y-2}{5}=(x+1)^{2}\qquad$ ...solve for $y$ by multiplying the entire expression with $5$.
$ y-2=5(x+1)^{2}\qquad$ ...add $2$ to each side.
$y=5(x+1)^{2}+2$
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, $h=-1,\ k=2$, so the vertex is $(-1,2)$
