(16x)^2+(9x)^2=24

Solution for (16x)^2+(9x)^2=24 equation:

(16x)^2+(9x)^2=24

We move all terms to the left:

(16x)^2+(9x)^2-(24)=0

We add all the numbers together, and all the variables

25x^2-24=0

a = 25; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·25·(-24)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*25}=\frac{0-20\sqrt{6}}{50}
=-\frac{20\sqrt{6}}{50}
=-\frac{2\sqrt{6}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*25}=\frac{0+20\sqrt{6}}{50}
=\frac{20\sqrt{6}}{50}
=\frac{2\sqrt{6}}{5}
$