(3x)^2+(4x)^2=625

Solution for (3x)^2+(4x)^2=625 equation:

(3x)^2+(4x)^2=625

We move all terms to the left:

(3x)^2+(4x)^2-(625)=0

We add all the numbers together, and all the variables

7x^2-625=0

a = 7; b = 0; c = -625;
Δ = b2-4ac
Δ = 02-4·7·(-625)
Δ = 17500
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:

$\sqrt{\Delta}=\sqrt{17500}=\sqrt{2500*7}=\sqrt{2500}*\sqrt{7}=50\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50\sqrt{7}}{2*7}=\frac{0-50\sqrt{7}}{14}
=-\frac{50\sqrt{7}}{14}
=-\frac{25\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50\sqrt{7}}{2*7}=\frac{0+50\sqrt{7}}{14}
=\frac{50\sqrt{7}}{14}
=\frac{25\sqrt{7}}{7}
$