2x^2+4x^2=320

Solution for 2x^2+4x^2=320 equation:

2x^2+4x^2=320

We move all terms to the left:

2x^2+4x^2-(320)=0

We add all the numbers together, and all the variables

6x^2-320=0

a = 6; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·6·(-320)
Δ = 7680
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:

$\sqrt{\Delta}=\sqrt{7680}=\sqrt{256*30}=\sqrt{256}*\sqrt{30}=16\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{30}}{2*6}=\frac{0-16\sqrt{30}}{12}
=-\frac{16\sqrt{30}}{12}
=-\frac{4\sqrt{30}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{30}}{2*6}=\frac{0+16\sqrt{30}}{12}
=\frac{16\sqrt{30}}{12}
=\frac{4\sqrt{30}}{3}
$