(9x)^2+(16x)^2=32

Solution for (9x)^2+(16x)^2=32 equation:

(9x)^2+(16x)^2=32

We move all terms to the left:

(9x)^2+(16x)^2-(32)=0

We add all the numbers together, and all the variables

25x^2-32=0

a = 25; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·25·(-32)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*25}=\frac{0-40\sqrt{2}}{50}
=-\frac{40\sqrt{2}}{50}
=-\frac{4\sqrt{2}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*25}=\frac{0+40\sqrt{2}}{50}
=\frac{40\sqrt{2}}{50}
=\frac{4\sqrt{2}}{5}
$