x^2+(3x)^2=576

Solution for x^2+(3x)^2=576 equation:

x^2+(3x)^2=576

We move all terms to the left:

x^2+(3x)^2-(576)=0

We add all the numbers together, and all the variables

4x^2-576=0

a = 4; b = 0; c = -576;
Δ = b2-4ac
Δ = 02-4·4·(-576)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96}{2*4}=\frac{-96}{8}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96}{2*4}=\frac{96}{8}
=12
$