-(2a)^2+2=-2

Solution for -(2a)^2+2=-2 equation:

-(2a)^2+2=-2

We move all terms to the left:

-(2a)^2+2-(-2)=0

We add all the numbers together, and all the variables

-2a^2+4=0

a = -2; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-2)·4
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-2}=\frac{0-4\sqrt{2}}{-4}
=-\frac{4\sqrt{2}}{-4}
=-\frac{\sqrt{2}}{-1}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-2}=\frac{0+4\sqrt{2}}{-4}
=\frac{4\sqrt{2}}{-4}
=\frac{\sqrt{2}}{-1}
$