X^2+1+(2x)^2+(2x)^2+1=50

Solution for X^2+1+(2x)^2+(2x)^2+1=50 equation:

X^2+1+(2X)^2+(2X)^2+1=50

We move all terms to the left:

X^2+1+(2X)^2+(2X)^2+1-(50)=0

We add all the numbers together, and all the variables

5X^2-48=0

a = 5; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·5·(-48)
Δ = 960
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:

$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*5}=\frac{0-8\sqrt{15}}{10}
=-\frac{8\sqrt{15}}{10}
=-\frac{4\sqrt{15}}{5}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*5}=\frac{0+8\sqrt{15}}{10}
=\frac{8\sqrt{15}}{10}
=\frac{4\sqrt{15}}{5}
$