x^2+(3x)^2=160

Solution for x^2+(3x)^2=160 equation:

x^2+(3x)^2=160

We move all terms to the left:

x^2+(3x)^2-(160)=0

We add all the numbers together, and all the variables

4x^2-160=0

a = 4; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·4·(-160)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*4}=\frac{0-16\sqrt{10}}{8}
=-\frac{16\sqrt{10}}{8}
=-2\sqrt{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*4}=\frac{0+16\sqrt{10}}{8}
=\frac{16\sqrt{10}}{8}
=2\sqrt{10}
$