((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 get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-288=0

a = 2; b = 0; c = -288;
Δ = b2-4ac
Δ = 02-4·2·(-288)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*2}=\frac{-48}{4}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*2}=\frac{48}{4}
=12
$