$2x^2-16x+28$ into standard form

Question

I think I'm just doing something stupid here, because I know it's not hard.

Here's what I did: $$y-28+{\_\_\_}=2x^2-16+{\_\_\_}$$ $$y-28+{\_\_\_}=2(x^2-8+{\_\_\_})$$ $$y-28+16= 2(x^2-8+16)$$ $$y-12=2(x-4)^2$$ $$y=2(x-4)^2+12$$

But then it ends up with different vertex than the original quadratic. What am I doing wrong here?

Answer 1

In the third line, you added 16 to the left side, and 32 to the right side.

Answer 2

Note that $$y=2(x-4)^2 - 4 \equiv 2x^2 - 16x + 28$$

Your mistake was in the third line, were you added $16$ to the LHS and $32$ to the RHS. $2 \cdot 16 \neq 16$.