(2x)^2=2^12

Solution for (2x)^2=2^12 equation:

(2x)^2=2^12

We move all terms to the left:

(2x)^2-(2^12)=0

We add all the numbers together, and all the variables

2x^2-4096=0

a = 2; b = 0; c = -4096;
Δ = b2-4ac
Δ = 02-4·2·(-4096)
Δ = 32768
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:

$\sqrt{\Delta}=\sqrt{32768}=\sqrt{16384*2}=\sqrt{16384}*\sqrt{2}=128\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128\sqrt{2}}{2*2}=\frac{0-128\sqrt{2}}{4}
=-\frac{128\sqrt{2}}{4}
=-32\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128\sqrt{2}}{2*2}=\frac{0+128\sqrt{2}}{4}
=\frac{128\sqrt{2}}{4}
=32\sqrt{2}
$