Htykuut

I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?

For example:
begin{align*}
f(x) & = x^2 + 6x + 11 \
& = (x^2 + 6x) + 11 \
& = (x^2 + 6x + mathbf{9}) + 11 - mathbf{9} \
& = (x+3)^2 + 2.
end{align*}
For positive you would add 9 inside the parenthesis and subtract 9 outside. However, I get the wrong answer when I do it with a negative coefficient. Do you do the same thing when there is a negative coefficient in front of $x^2$ or is it the other way around? (subtract inside parenthesis and add outside of parenthesis).

Here is an example of a negative coefficient:
$$f(x) = -3x^2 + 5x + 1.$$
I tried to solve this and entered in my answer online but it was wrong.

$
begin{split}
f(x) &= -3(x^2-5x/3 - 1/3)\
&= -3( x^2 - 5x/3 + (5/6)^2 - (5/6)^2 - 1/3)\
&= -3( (x - 5/6)^2 - 25/36 - 12/36) \
&= -3(x - 5/6)^2 -3(- 25/36 - 12/36) \
&= -3(x - 5/6)^2 + 37/12.
end{split}$

$$(2ax+b)^2 = 4a^2x^2 + 4abx + b^2$$

So, multiplying by $4a$, or in this case $-4a$, will avoid fractions until the last step.

begin{align}
f(x) &= -3x^2 + 5x + 1 \
-12f(x) &= 36x^2 -60x - 12 \
-12f(x) &= 36x^2 - 60x + 25 - 25 - 12 \
-12f(x) &= (6x-5)^2 - 37 & left(b = frac{4ab}{4a} = dfrac{-60}{12}=-5 right)\
f(x) &= -dfrac{1}{12}(6x-5)^2 + dfrac{37}{12}
end{align}