x^2+10+x^2-98=180

Solution for x^2+10+x^2-98=180 equation:

x^2+10+x^2-98=180

We move all terms to the left:

x^2+10+x^2-98-(180)=0

We add all the numbers together, and all the variables

2x^2-268=0

a = 2; b = 0; c = -268;
Δ = b2-4ac
Δ = 02-4·2·(-268)
Δ = 2144
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:

$\sqrt{\Delta}=\sqrt{2144}=\sqrt{16*134}=\sqrt{16}*\sqrt{134}=4\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{134}}{2*2}=\frac{0-4\sqrt{134}}{4}
=-\frac{4\sqrt{134}}{4}
=-\sqrt{134}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{134}}{2*2}=\frac{0+4\sqrt{134}}{4}
=\frac{4\sqrt{134}}{4}
=\sqrt{134}
$