-x2+40x+500=0

Solution for -x2+40x+500=0 equation:

-x2+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{-100}{-2}
=+50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+60}{2*-1}=\frac{20}{-2}
=-10
$