2x^2+x^2=320^2

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

2x^2+x^2=320^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

3x^2-102400=0

a = 3; b = 0; c = -102400;
Δ = b2-4ac
Δ = 02-4·3·(-102400)
Δ = 1228800
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:

$\sqrt{\Delta}=\sqrt{1228800}=\sqrt{409600*3}=\sqrt{409600}*\sqrt{3}=640\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-640\sqrt{3}}{2*3}=\frac{0-640\sqrt{3}}{6}
=-\frac{640\sqrt{3}}{6}
=-\frac{320\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+640\sqrt{3}}{2*3}=\frac{0+640\sqrt{3}}{6}
=\frac{640\sqrt{3}}{6}
=\frac{320\sqrt{3}}{3}
$