(2x)^2=128

Solution for (2x)^2=128 equation:

(2x)^2=128

We move all terms to the left:

(2x)^2-(128)=0

a = 2; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·2·(-128)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*2}=\frac{32}{4}
=8
$