(2x)^2+4x^2=120

Solution for (2x)^2+4x^2=120 equation:

(2x)^2+4x^2=120

We move all terms to the left:

(2x)^2+4x^2-(120)=0

We add all the numbers together, and all the variables

6x^2-120=0

a = 6; b = 0; c = -120;
Δ = b2-4ac
Δ = 02-4·6·(-120)
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{5}}{2*6}=\frac{0-24\sqrt{5}}{12}
=-\frac{24\sqrt{5}}{12}
=-2\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{5}}{2*6}=\frac{0+24\sqrt{5}}{12}
=\frac{24\sqrt{5}}{12}
=2\sqrt{5}
$