x^2+(2x)^2=125

Solution for x^2+(2x)^2=125 equation:

x^2+(2x)^2=125

We move all terms to the left:

x^2+(2x)^2-(125)=0

We add all the numbers together, and all the variables

3x^2-125=0

a = 3; b = 0; c = -125;
Δ = b2-4ac
Δ = 02-4·3·(-125)
Δ = 1500
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:

$\sqrt{\Delta}=\sqrt{1500}=\sqrt{100*15}=\sqrt{100}*\sqrt{15}=10\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{15}}{2*3}=\frac{0-10\sqrt{15}}{6}
=-\frac{10\sqrt{15}}{6}
=-\frac{5\sqrt{15}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{15}}{2*3}=\frac{0+10\sqrt{15}}{6}
=\frac{10\sqrt{15}}{6}
=\frac{5\sqrt{15}}{3}
$