2x^2-4x+8=x^2+x+2

Solution for 2x^2-4x+8=x^2+x+2 equation:

2x^2-4x+8=x^2+x+2

We move all terms to the left:

2x^2-4x+8-(x^2+x+2)=0

We get rid of parentheses

2x^2-x^2-4x-x-2+8=0

We add all the numbers together, and all the variables

x^2-5x+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$