x^2+2x-999=0

Solution for x^2+2x-999=0 equation:

x^2+2x-999=0

a = 1; b = 2; c = -999;
Δ = b2-4ac
Δ = 22-4·1·(-999)
Δ = 4000
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:

$\sqrt{\Delta}=\sqrt{4000}=\sqrt{400*10}=\sqrt{400}*\sqrt{10}=20\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-20\sqrt{10}}{2*1}=\frac{-2-20\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+20\sqrt{10}}{2*1}=\frac{-2+20\sqrt{10}}{2}
$