2x^2-8=x^2+1

Solution for 2x^2-8=x^2+1 equation:

2x^2-8=x^2+1

We move all terms to the left:

2x^2-8-(x^2+1)=0

We get rid of parentheses

2x^2-x^2-1-8=0

We add all the numbers together, and all the variables

x^2-9=0

a = 1; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·1·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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