-4x^2+16x-20=-36

Solution for -4x^2+16x-20=-36 equation:

-4x^2+16x-20=-36

We move all terms to the left:

-4x^2+16x-20-(-36)=0

We add all the numbers together, and all the variables

-4x^2+16x+16=0

a = -4; b = 16; c = +16;
Δ = b2-4ac
Δ = 162-4·(-4)·16
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*-4}=\frac{-16-16\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*-4}=\frac{-16+16\sqrt{2}}{-8}
$