(2x)^2+x^2=400

Solution for (2x)^2+x^2=400 equation:

(2x)^2+x^2=400

We move all terms to the left:

(2x)^2+x^2-(400)=0

We add all the numbers together, and all the variables

3x^2-400=0

a = 3; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·3·(-400)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*3}=\frac{0-40\sqrt{3}}{6}
=-\frac{40\sqrt{3}}{6}
=-\frac{20\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*3}=\frac{0+40\sqrt{3}}{6}
=\frac{40\sqrt{3}}{6}
=\frac{20\sqrt{3}}{3}
$