(2x+8)(4x-16)=128

Solution for (2x+8)(4x-16)=128 equation:

(2x+8)(4x-16)=128

We move all terms to the left:

(2x+8)(4x-16)-(128)=0

We multiply parentheses ..

(+8x^2-32x+32x-128)-128=0

We get rid of parentheses

8x^2-32x+32x-128-128=0

We add all the numbers together, and all the variables

8x^2-256=0

a = 8; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·8·(-256)
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{2}}{2*8}=\frac{0-64\sqrt{2}}{16}
=-\frac{64\sqrt{2}}{16}
=-4\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{2}}{2*8}=\frac{0+64\sqrt{2}}{16}
=\frac{64\sqrt{2}}{16}
=4\sqrt{2}
$