n(n+1)(n-1)=1000

Solution for n(n+1)(n-1)=1000 equation:

n(n+1)(n-1)=1000

We move all terms to the left:

n(n+1)(n-1)-(1000)=0

We use the square of the difference formula

n^2-1-1000=0

We add all the numbers together, and all the variables

n^2-1001=0

a = 1; b = 0; c = -1001;
Δ = b2-4ac
Δ = 02-4·1·(-1001)
Δ = 4004
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:

$\sqrt{\Delta}=\sqrt{4004}=\sqrt{4*1001}=\sqrt{4}*\sqrt{1001}=2\sqrt{1001}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1001}}{2*1}=\frac{0-2\sqrt{1001}}{2}
=-\frac{2\sqrt{1001}}{2}
=-\sqrt{1001}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1001}}{2*1}=\frac{0+2\sqrt{1001}}{2}
=\frac{2\sqrt{1001}}{2}
=\sqrt{1001}
$