(5x)^2+(6x)^2=122^2

Solution for (5x)^2+(6x)^2=122^2 equation:

(5x)^2+(6x)^2=122^2

We move all terms to the left:

(5x)^2+(6x)^2-(122^2)=0

We add all the numbers together, and all the variables

11x^2-14884=0

a = 11; b = 0; c = -14884;
Δ = b2-4ac
Δ = 02-4·11·(-14884)
Δ = 654896
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:

$\sqrt{\Delta}=\sqrt{654896}=\sqrt{59536*11}=\sqrt{59536}*\sqrt{11}=244\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-244\sqrt{11}}{2*11}=\frac{0-244\sqrt{11}}{22}
=-\frac{244\sqrt{11}}{22}
=-\frac{122\sqrt{11}}{11}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+244\sqrt{11}}{2*11}=\frac{0+244\sqrt{11}}{22}
=\frac{244\sqrt{11}}{22}
=\frac{122\sqrt{11}}{11}
$