(16x)^2+(9x)^2=140^2

Solution for (16x)^2+(9x)^2=140^2 equation:

(16x)^2+(9x)^2=140^2

We move all terms to the left:

(16x)^2+(9x)^2-(140^2)=0

We add all the numbers together, and all the variables

25x^2-19600=0

a = 25; b = 0; c = -19600;
Δ = b2-4ac
Δ = 02-4·25·(-19600)
Δ = 1960000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1960000}=1400$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-1400}{2*25}=\frac{-1400}{50}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+1400}{2*25}=\frac{1400}{50}
=28
$