x^2+(2x)^2=40

Solution for x^2+(2x)^2=40 equation:

x^2+(2x)^2=40

We move all terms to the left:

x^2+(2x)^2-(40)=0

We add all the numbers together, and all the variables

3x^2-40=0

a = 3; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·3·(-40)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*3}=\frac{0-4\sqrt{30}}{6}
=-\frac{4\sqrt{30}}{6}
=-\frac{2\sqrt{30}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*3}=\frac{0+4\sqrt{30}}{6}
=\frac{4\sqrt{30}}{6}
=\frac{2\sqrt{30}}{3}
$