(16x)^2+(9x)^2=27^2

Solution for (16x)^2+(9x)^2=27^2 equation:

(16x)^2+(9x)^2=27^2

We move all terms to the left:

(16x)^2+(9x)^2-(27^2)=0

We add all the numbers together, and all the variables

25x^2-729=0

a = 25; b = 0; c = -729;
Δ = b2-4ac
Δ = 02-4·25·(-729)
Δ = 72900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{72900}=270$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-270}{2*25}=\frac{-270}{50}
=-5+2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+270}{2*25}=\frac{270}{50}
=5+2/5
$