(9x)^2+(16x)^2=55

Solution for (9x)^2+(16x)^2=55 equation:

(9x)^2+(16x)^2=55

We move all terms to the left:

(9x)^2+(16x)^2-(55)=0

We add all the numbers together, and all the variables

25x^2-55=0

a = 25; b = 0; c = -55;
Δ = b2-4ac
Δ = 02-4·25·(-55)
Δ = 5500
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:

$\sqrt{\Delta}=\sqrt{5500}=\sqrt{100*55}=\sqrt{100}*\sqrt{55}=10\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{55}}{2*25}=\frac{0-10\sqrt{55}}{50}
=-\frac{10\sqrt{55}}{50}
=-\frac{\sqrt{55}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{55}}{2*25}=\frac{0+10\sqrt{55}}{50}
=\frac{10\sqrt{55}}{50}
=\frac{\sqrt{55}}{5}
$