(2x)^2+x^2=320

Solution for (2x)^2+x^2=320 equation:

(2x)^2+x^2=320

We move all terms to the left:

(2x)^2+x^2-(320)=0

We add all the numbers together, and all the variables

3x^2-320=0

a = 3; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·3·(-320)
Δ = 3840
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:

$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{15}}{2*3}=\frac{0-16\sqrt{15}}{6}
=-\frac{16\sqrt{15}}{6}
=-\frac{8\sqrt{15}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{15}}{2*3}=\frac{0+16\sqrt{15}}{6}
=\frac{16\sqrt{15}}{6}
=\frac{8\sqrt{15}}{3}
$