x^2-4x+4-4x^2+12x+9=-80

Solution for x^2-4x+4-4x^2+12x+9=-80 equation:

x^2-4x+4-4x^2+12x+9=-80

We move all terms to the left:

x^2-4x+4-4x^2+12x+9-(-80)=0

We add all the numbers together, and all the variables

-3x^2+8x+93=0

a = -3; b = 8; c = +93;
Δ = b2-4ac
Δ = 82-4·(-3)·93
Δ = 1180
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:

$\sqrt{\Delta}=\sqrt{1180}=\sqrt{4*295}=\sqrt{4}*\sqrt{295}=2\sqrt{295}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{295}}{2*-3}=\frac{-8-2\sqrt{295}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{295}}{2*-3}=\frac{-8+2\sqrt{295}}{-6}
$