2001x^2+2x-1999=0

Solution for 2001x^2+2x-1999=0 equation:

2001x^2+2x-1999=0

a = 2001; b = 2; c = -1999;
Δ = b2-4ac
Δ = 22-4·2001·(-1999)
Δ = 16000000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16000000}=4000$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4000}{2*2001}=\frac{-4002}{4002}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4000}{2*2001}=\frac{3998}{4002}
=1999/2001
$