x^2+(3x)^2=400

Solution for x^2+(3x)^2=400 equation:

x^2+(3x)^2=400

We move all terms to the left:

x^2+(3x)^2-(400)=0

We add all the numbers together, and all the variables

4x^2-400=0

a = 4; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·4·(-400)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*4}=\frac{-80}{8}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*4}=\frac{80}{8}
=10
$