x^2+80x-500=0

Solution for x^2+80x-500=0 equation:

x^2+80x-500=0

a = 1; b = 80; c = -500;
Δ = b2-4ac
Δ = 802-4·1·(-500)
Δ = 8400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8400}=\sqrt{400*21}=\sqrt{400}*\sqrt{21}=20\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-20\sqrt{21}}{2*1}=\frac{-80-20\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+20\sqrt{21}}{2*1}=\frac{-80+20\sqrt{21}}{2}
$