N(n+1)(n-1)=990

Solution for N(n+1)(n-1)=990 equation:

(N+1)(N-1)=990

We move all terms to the left:

(N+1)(N-1)-(990)=0

We use the square of the difference formula

N^2-1-990=0

We add all the numbers together, and all the variables

N^2-991=0

a = 1; b = 0; c = -991;
Δ = b2-4ac
Δ = 02-4·1·(-991)
Δ = 3964
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:

$\sqrt{\Delta}=\sqrt{3964}=\sqrt{4*991}=\sqrt{4}*\sqrt{991}=2\sqrt{991}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{991}}{2*1}=\frac{0-2\sqrt{991}}{2}
=-\frac{2\sqrt{991}}{2}
=-\sqrt{991}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{991}}{2*1}=\frac{0+2\sqrt{991}}{2}
=\frac{2\sqrt{991}}{2}
=\sqrt{991}
$