(2x)^2+(3x)^2=64

Solution for (2x)^2+(3x)^2=64 equation:

(2x)^2+(3x)^2=64

We move all terms to the left:

(2x)^2+(3x)^2-(64)=0

We add all the numbers together, and all the variables

5x^2-64=0

a = 5; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·5·(-64)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*5}=\frac{0-16\sqrt{5}}{10}
=-\frac{16\sqrt{5}}{10}
=-\frac{8\sqrt{5}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*5}=\frac{0+16\sqrt{5}}{10}
=\frac{16\sqrt{5}}{10}
=\frac{8\sqrt{5}}{5}
$