-y^2-4+12=y

Solution for -y^2-4+12=y equation:

-y^2-4+12=y

We move all terms to the left:

-y^2-4+12-(y)=0

We add all the numbers together, and all the variables

-1y^2-1y+8=0

a = -1; b = -1; c = +8;
Δ = b2-4ac
Δ = -12-4·(-1)·8
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{33}}{2*-1}=\frac{1-\sqrt{33}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{33}}{2*-1}=\frac{1+\sqrt{33}}{-2}
$