Rewriting $2x^2-12x+11$ by completing the square I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?

An alternative way to see how this all works is to note that completing the square doesn't change the value of the expression. What you are essentially doing is adding a cleverly-written $0$ to it so that it becomes easier to work with.
Observe, $$\begin{align}2x^2-12x+11 & = 2x^2 - 12x \qquad\qquad\qquad\qquad\qquad\quad +11 \tag{Leave space}\\ &= 2[x^2 - 6x]\qquad\qquad\qquad\qquad\qquad\;\; +11 \tag{Factor}\\&= 2[x^2 - 6x +\underbrace{\left({-6\over2}\right)^2 - \left({-6\over2}\right)^2}_\text{cleverly-written zero}] + 11 \\ &= 2[\underbrace{x^2 - 6x + 9}_{(x-3)^2} - 9] + 11\\ &= 2[(x-3)^2 - 9] + 11\\ &= 2(x-3)^2 - 18 + 11\\ &= 2(x-3)^2 - 7.\end{align}$$
