x^2+4x-500=0

Solution for x^2+4x-500=0 equation:

x^2+4x-500=0

a = 1; b = 4; c = -500;
Δ = b2-4ac
Δ = 42-4·1·(-500)
Δ = 2016
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:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{14}}{2*1}=\frac{-4-12\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{14}}{2*1}=\frac{-4+12\sqrt{14}}{2}
$