x^2+(2x)^2=25^2

Solution for x^2+(2x)^2=25^2 equation:

x^2+(2x)^2=25^2

We move all terms to the left:

x^2+(2x)^2-(25^2)=0

We add all the numbers together, and all the variables

3x^2-625=0

a = 3; b = 0; c = -625;
Δ = b2-4ac
Δ = 02-4·3·(-625)
Δ = 7500
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:

$\sqrt{\Delta}=\sqrt{7500}=\sqrt{2500*3}=\sqrt{2500}*\sqrt{3}=50\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50\sqrt{3}}{2*3}=\frac{0-50\sqrt{3}}{6}
=-\frac{50\sqrt{3}}{6}
=-\frac{25\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50\sqrt{3}}{2*3}=\frac{0+50\sqrt{3}}{6}
=\frac{50\sqrt{3}}{6}
=\frac{25\sqrt{3}}{3}
$