(3x)^2+(2x)^2=117

Solution for (3x)^2+(2x)^2=117 equation:

(3x)^2+(2x)^2=117

We move all terms to the left:

(3x)^2+(2x)^2-(117)=0

We add all the numbers together, and all the variables

5x^2-117=0

a = 5; b = 0; c = -117;
Δ = b2-4ac
Δ = 02-4·5·(-117)
Δ = 2340
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:

$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{65}}{2*5}=\frac{0-6\sqrt{65}}{10}
=-\frac{6\sqrt{65}}{10}
=-\frac{3\sqrt{65}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{65}}{2*5}=\frac{0+6\sqrt{65}}{10}
=\frac{6\sqrt{65}}{10}
=\frac{3\sqrt{65}}{5}
$