How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$ 
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?

Method 1:
$y = -4x^2 -2x -4$
$\frac {y}{-4} = x^2 + \frac 12x + 1$
$-\frac {y}{4}- 1 = x^2 + 2*\frac 14 x$
$-1-\frac y4  +(\frac 14)^2= x^2 + 2*\frac 14 x+ (\frac 14)^2$
$-1+\frac 1{16}-\frac y4  = (x + \frac 14)^2$
$(x+  \frac 14)^2 = -\frac {15}{16} - \frac y4 = \frac {-15 - 4y}{16}$
$x + \frac 14 = \pm \sqrt {\frac {-15-4y}{16} }= \frac {\sqrt {-15 - 4y}}4$
$x = \frac {\sqrt {-15 - 4y}}4 -\frac 14 = \frac {\sqrt{-15 - 4y} - 1}4$
Method 2:
$y = -4x^2 - 2x -4 \iff y+4 = -4x^2 -2x$ 
We want $-4x^2 - 2x$ to be the first to summands of $(mx + n)^2 = m^2x + 2mnx + n^2$.  But $m^2$ is positive while $-4$ is negative.
So we multiple everything by $-1$.
$y+4 = -4x^2 -2x \iff -y - 4 = 4x^2 + 2x$.  Now we want $4 = m^2$ and $2 = 2mn$.
Or in other words $m = \pm \sqrt 4$ and $n = \frac 1m$. Or in other words $m = \pm 2$ and $n = \frac 12$.  
$-y -4 = 4x^2 + 2x = m^2x^2 + 2mnx \iff$
$-y-4 + n^2 = -y -4 + (\frac 12)^2= 4x^2 + 2x + (\frac 12)^2 = m^2x^2 + 2mnx +n^2 = (mx + n)^2 = (2x + \frac 12)^2$
$\iff -y -4 +\frac 14 = -y -\frac {15}4 = (2x + \frac 12)^2$
$\iff 2x +\frac 12 =\pm \sqrt {-y -\frac {15}{4}}$
$\iff 2x =-\frac 12 \pm \sqrt {-y-\frac {15}4}=-\frac 12 \pm \frac {\sqrt{-4y-15}}2 = \frac {-1 \pm\sqrt {-4y - 15}}2$
$\iff x = \frac {-1 \pm \sqrt{-4y -15}}4$ 
