2x^2+20x-500=0

Solution for 2x^2+20x-500=0 equation:

2x^2+20x-500=0

a = 2; b = 20; c = -500;
Δ = b2-4ac
Δ = 202-4·2·(-500)
Δ = 4400
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:

$\sqrt{\Delta}=\sqrt{4400}=\sqrt{400*11}=\sqrt{400}*\sqrt{11}=20\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{11}}{2*2}=\frac{-20-20\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{11}}{2*2}=\frac{-20+20\sqrt{11}}{4}
$