X^2+(5x)^2=104

Solution for X^2+(5x)^2=104 equation:

X^2+(5X)^2=104

We move all terms to the left:

X^2+(5X)^2-(104)=0

We add all the numbers together, and all the variables

6X^2-104=0

a = 6; b = 0; c = -104;
Δ = b2-4ac
Δ = 02-4·6·(-104)
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{39}}{2*6}=\frac{0-8\sqrt{39}}{12}
=-\frac{8\sqrt{39}}{12}
=-\frac{2\sqrt{39}}{3}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{39}}{2*6}=\frac{0+8\sqrt{39}}{12}
=\frac{8\sqrt{39}}{12}
=\frac{2\sqrt{39}}{3}
$