2x^2-3x-8=-x^2+2x

Solution for 2x^2-3x-8=-x^2+2x equation:

2x^2-3x-8=-x^2+2x

We move all terms to the left:

2x^2-3x-8-(-x^2+2x)=0

We get rid of parentheses

2x^2+x^2-2x-3x-8=0

We add all the numbers together, and all the variables

3x^2-5x-8=0

a = 3; b = -5; c = -8;
Δ = b2-4ac
Δ = -52-4·3·(-8)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*3}=\frac{16}{6}
=2+2/3
$