X^2+(x^2+1)+(x^2+4)-50=0

Solution for X^2+(x^2+1)+(x^2+4)-50=0 equation:

X^2+(X^2+1)+(X^2+4)-50=0

We get rid of parentheses

X^2+X^2+X^2+1+4-50=0

We add all the numbers together, and all the variables

3X^2-45=0

a = 3; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·3·(-45)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*3}=\frac{0-6\sqrt{15}}{6}
=-\frac{6\sqrt{15}}{6}
=-\sqrt{15}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*3}=\frac{0+6\sqrt{15}}{6}
=\frac{6\sqrt{15}}{6}
=\sqrt{15}
$