Answer
$y=2\left(x-\frac{3}{2}\right)^2-\frac{11}{2}$
Work Step by Step
\begin{align*}
&y=(2x^2-6x)-1 &\text{Group the terms with variables together.}\\
&y=2(x^2-3x)-1 &\text{Factor out $2$.}\\
\end{align*}
Recall:
To complete the square of $y=a(x^2+bx)+c$, you add $\left(\dfrac{b}{2}\right)^2$ inside the parentheses and subtract $a\cdot \left(\dfrac{b}{2}\right)^2$ outside to obtain $y=a\left[x^2+bx +\color{blue}{\left(\dfrac{b}{2}\right)^2}\right]+c-\color{red}{a\left(\dfrac{b}{2}\right)^2}$
\begin{align*}
&y=2\left[x^2-3x+\color{blue}{\left(\frac{-3}{2}\right)^2}\right]-1-\color{red}{2\left(\frac{-3}{2}\right)^2} &\text{Complete the square.}\\\\
&y=2\left[x^2-3x+\frac{9}{4}\right]-1-2\left(\frac{9}{4}\right) &\text{Simplify.}\\\\
&y=2\left(x-\frac{3}{2}\right)^2-1-\frac{9}{2}\\\\
&y=2\left(x-\frac{3}{2}\right)^2-\frac{2}{2}-\frac{9}{2}\\\\
&y=2\left(x-\frac{3}{2}\right)^2-\frac{11}{2}\\\\
\end{align*}