x^2+(4x)^2=68

Solution for x^2+(4x)^2=68 equation:

x^2+(4x)^2=68

We move all terms to the left:

x^2+(4x)^2-(68)=0

We add all the numbers together, and all the variables

5x^2-68=0

a = 5; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·5·(-68)
Δ = 1360
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:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{85}}{2*5}=\frac{0-4\sqrt{85}}{10}
=-\frac{4\sqrt{85}}{10}
=-\frac{2\sqrt{85}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{85}}{2*5}=\frac{0+4\sqrt{85}}{10}
=\frac{4\sqrt{85}}{10}
=\frac{2\sqrt{85}}{5}
$