20=x^2+(2x)^2

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

20=x^2+(2x)^2

We move all terms to the left:

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

We get rid of parentheses

-x^2-2x^2+20=0

We add all the numbers together, and all the variables

-3x^2+20=0

a = -3; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-3)·20
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-3}=\frac{0-4\sqrt{15}}{-6}
=-\frac{4\sqrt{15}}{-6}
=-\frac{2\sqrt{15}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-3}=\frac{0+4\sqrt{15}}{-6}
=\frac{4\sqrt{15}}{-6}
=\frac{2\sqrt{15}}{-3}
$