(x^2+4x-3)-2(x+4)(-2x+2)=0

Solution for (x^2+4x-3)-2(x+4)(-2x+2)=0 equation:

(x^2+4x-3)-2(x+4)(-2x+2)=0

We get rid of parentheses

x^2+4x-2(x+4)(-2x+2)-3=0

We multiply parentheses ..

x^2-2(-2x^2+2x-8x+8)+4x-3=0

We multiply parentheses

x^2+4x^2-4x+16x+4x-16-3=0

We add all the numbers together, and all the variables

5x^2+16x-19=0

a = 5; b = 16; c = -19;
Δ = b2-4ac
Δ = 162-4·5·(-19)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{159}}{2*5}=\frac{-16-2\sqrt{159}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{159}}{2*5}=\frac{-16+2\sqrt{159}}{10}
$