Answer
Vertex: $(1,-4)$
Zeros: $x= \pm \sqrt{2 }+1$
Work Step by Step
We have $y=2 x^2-4 x-2=2\left(x^2-2 x-1\right)$. Completing the square, we get
$$
\begin{aligned}
y & =2\left(x^2-2 x+1^2-1^2-1\right) \\
& =2(x-1)^2-4
\end{aligned}
$$
The vertex is at $(1,-4)$. The zeros occur when $y=0$, so
$$
\begin{aligned}
& 0=2(x-1)^2-4 \\
& 2=(x-1)^2 \\
& x= \pm \sqrt{2}+1
\end{aligned}
$$
