(2x)^2=20

Solution for (2x)^2=20 equation:

(2x)^2=20

We move all terms to the left:

(2x)^2-(20)=0

a = 2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·2·(-20)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*2}=\frac{0-4\sqrt{10}}{4}
=-\frac{4\sqrt{10}}{4}
=-\sqrt{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*2}=\frac{0+4\sqrt{10}}{4}
=\frac{4\sqrt{10}}{4}
=\sqrt{10}
$