(3x)^2+(4x)^2=100

Solution for (3x)^2+(4x)^2=100 equation:

(3x)^2+(4x)^2=100

We move all terms to the left:

(3x)^2+(4x)^2-(100)=0

We add all the numbers together, and all the variables

7x^2-100=0

a = 7; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·7·(-100)
Δ = 2800
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:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{7}}{2*7}=\frac{0-20\sqrt{7}}{14}
=-\frac{20\sqrt{7}}{14}
=-\frac{10\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{7}}{2*7}=\frac{0+20\sqrt{7}}{14}
=\frac{20\sqrt{7}}{14}
=\frac{10\sqrt{7}}{7}
$