0=1x^2+2x-8

Solution for 0=1x^2+2x-8 equation:

0=1x^2+2x-8

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
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{-(-2)-6}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2}
=-4
$