(1+n)n=2000

Solution for (1+n)n=2000 equation:

(1+n)n=2000

We move all terms to the left:

(1+n)n-(2000)=0

We add all the numbers together, and all the variables

(n+1)n-2000=0

We multiply parentheses

n^2+n-2000=0

a = 1; b = 1; c = -2000;
Δ = b2-4ac
Δ = 12-4·1·(-2000)
Δ = 8001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8001}=\sqrt{9*889}=\sqrt{9}*\sqrt{889}=3\sqrt{889}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{889}}{2*1}=\frac{-1-3\sqrt{889}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{889}}{2*1}=\frac{-1+3\sqrt{889}}{2}
$