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

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

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

We move all terms to the left:

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

We multiply parentheses

4x^2-(2x^2+8x)-16=0

We get rid of parentheses

4x^2-2x^2-8x-16=0

We add all the numbers together, and all the variables

2x^2-8x-16=0

a = 2; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·2·(-16)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*2}=\frac{8-8\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*2}=\frac{8+8\sqrt{3}}{4}
$