0=x^2+40x-500

Solution for 0=x^2+40x-500 equation:

0=x^2+40x-500

We move all terms to the left:

0-(x^2+40x-500)=0

We add all the numbers together, and all the variables

-(x^2+40x-500)=0

We get rid of parentheses

-x^2-40x+500=0

We add all the numbers together, and all the variables

-1x^2-40x+500=0

a = -1; b = -40; c = +500;
Δ = b2-4ac
Δ = -402-4·(-1)·500
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-60}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+60}{2*-1}=\frac{100}{-2}
=-50
$