2x^2+5x-500=0

Solution for 2x^2+5x-500=0 equation:

2x^2+5x-500=0

a = 2; b = 5; c = -500;
Δ = b2-4ac
Δ = 52-4·2·(-500)
Δ = 4025
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:

$\sqrt{\Delta}=\sqrt{4025}=\sqrt{25*161}=\sqrt{25}*\sqrt{161}=5\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{161}}{2*2}=\frac{-5-5\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{161}}{2*2}=\frac{-5+5\sqrt{161}}{4}
$