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

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

X^2+(2X)^2=40^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

3X^2-1600=0

a = 3; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·3·(-1600)
Δ = 19200
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:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{3}}{2*3}=\frac{0-80\sqrt{3}}{6}
=-\frac{80\sqrt{3}}{6}
=-\frac{40\sqrt{3}}{3}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{3}}{2*3}=\frac{0+80\sqrt{3}}{6}
=\frac{80\sqrt{3}}{6}
=\frac{40\sqrt{3}}{3}
$