(3x)^2+(5x)^2=64

Solution for (3x)^2+(5x)^2=64 equation:

(3x)^2+(5x)^2=64

We move all terms to the left:

(3x)^2+(5x)^2-(64)=0

We add all the numbers together, and all the variables

8x^2-64=0

a = 8; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·8·(-64)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*8}=\frac{0-32\sqrt{2}}{16}
=-\frac{32\sqrt{2}}{16}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*8}=\frac{0+32\sqrt{2}}{16}
=\frac{32\sqrt{2}}{16}
=2\sqrt{2}
$