x^2+(3x)^2=40^2

Solution for x^2+(3x)^2=40^2 equation:

x^2+(3x)^2=40^2

We move all terms to the left:

x^2+(3x)^2-(40^2)=0

We add all the numbers together, and all the variables

4x^2-1600=0

a = 4; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·4·(-1600)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160}{2*4}=\frac{-160}{8}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160}{2*4}=\frac{160}{8}
=20
$