40=x^2+2x-8

Solution for 40=x^2+2x-8 equation:

40=x^2+2x-8

We move all terms to the left:

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

We get rid of parentheses

-x^2-2x+8+40=0

We add all the numbers together, and all the variables

-1x^2-2x+48=0

a = -1; b = -2; c = +48;
Δ = b2-4ac
Δ = -22-4·(-1)·48
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-1}=\frac{16}{-2}
=-8
$