2x^2+9x-8=x^2-8

Solution for 2x^2+9x-8=x^2-8 equation:

2x^2+9x-8=x^2-8

We move all terms to the left:

2x^2+9x-8-(x^2-8)=0

We get rid of parentheses

2x^2-x^2+9x+8-8=0

We add all the numbers together, and all the variables

x^2+9x=0

a = 1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*1}=\frac{0}{2}
=0
$