(3x)^2+(2x)^2=13^2

Solution for (3x)^2+(2x)^2=13^2 equation:

(3x)^2+(2x)^2=13^2

We move all terms to the left:

(3x)^2+(2x)^2-(13^2)=0

We add all the numbers together, and all the variables

5x^2-169=0

a = 5; b = 0; c = -169;
Δ = b2-4ac
Δ = 02-4·5·(-169)
Δ = 3380
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:

$\sqrt{\Delta}=\sqrt{3380}=\sqrt{676*5}=\sqrt{676}*\sqrt{5}=26\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-26\sqrt{5}}{2*5}=\frac{0-26\sqrt{5}}{10}
=-\frac{26\sqrt{5}}{10}
=-\frac{13\sqrt{5}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+26\sqrt{5}}{2*5}=\frac{0+26\sqrt{5}}{10}
=\frac{26\sqrt{5}}{10}
=\frac{13\sqrt{5}}{5}
$