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

Solution for 4(x+2)(x-2)=(2x)^2+8x equation:

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

We move all terms to the left:

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

We use the square of the difference formula

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

We get rid of parentheses

x^2-2x^2-8x-4=0

We add all the numbers together, and all the variables

-1x^2-8x-4=0

a = -1; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·(-1)·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*-1}=\frac{8-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*-1}=\frac{8+4\sqrt{3}}{-2}
$