x^2+2x-2000=0

Solution for x^2+2x-2000=0 equation:

x^2+2x-2000=0

a = 1; b = 2; c = -2000;
Δ = b2-4ac
Δ = 22-4·1·(-2000)
Δ = 8004
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:

$\sqrt{\Delta}=\sqrt{8004}=\sqrt{4*2001}=\sqrt{4}*\sqrt{2001}=2\sqrt{2001}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2001}}{2*1}=\frac{-2-2\sqrt{2001}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2001}}{2*1}=\frac{-2+2\sqrt{2001}}{2}
$