(2x)^2=16

Solution for (2x)^2=16 equation:

(2x)^2=16

We move all terms to the left:

(2x)^2-(16)=0

a = 2; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·2·(-16)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*2}=\frac{0-8\sqrt{2}}{4}
=-\frac{8\sqrt{2}}{4}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*2}=\frac{0+8\sqrt{2}}{4}
=\frac{8\sqrt{2}}{4}
=2\sqrt{2}
$