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 add all the numbers together, and all the variables

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

We use the square of the difference formula

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

We get rid of parentheses

-2x^2+x^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}
$