2x(4x+2x)=128

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

2x(4x+2x)=128

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x(+6x)-128=0

We multiply parentheses

12x^2-128=0

a = 12; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·12·(-128)
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{6}}{2*12}=\frac{0-32\sqrt{6}}{24}
=-\frac{32\sqrt{6}}{24}
=-\frac{4\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{6}}{2*12}=\frac{0+32\sqrt{6}}{24}
=\frac{32\sqrt{6}}{24}
=\frac{4\sqrt{6}}{3}
$