2x^2+8x=x^2+2x-8.

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

2x^2+8x=x^2+2x-8.

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+6x+8=0

a = 1; b = 6; c = +8;
Δ = b2-4ac
Δ = 62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*1}=\frac{-4}{2}
=-2
$