x2+(5x)2-26=0

Solution for x2+(5x)2-26=0 equation:

x2+(5x)2-26=0

We add all the numbers together, and all the variables

6x^2-26=0

a = 6; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·6·(-26)
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{39}}{2*6}=\frac{0-4\sqrt{39}}{12}
=-\frac{4\sqrt{39}}{12}
=-\frac{\sqrt{39}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{39}}{2*6}=\frac{0+4\sqrt{39}}{12}
=\frac{4\sqrt{39}}{12}
=\frac{\sqrt{39}}{3}
$