N(n+1)(n+2)=120

Solution for N(n+1)(n+2)=120 equation:

(N+1)(N+2)=120

We move all terms to the left:

(N+1)(N+2)-(120)=0

We multiply parentheses ..

(+N^2+2N+N+2)-120=0

We get rid of parentheses

N^2+2N+N+2-120=0

We add all the numbers together, and all the variables

N^2+3N-118=0

a = 1; b = 3; c = -118;
Δ = b2-4ac
Δ = 32-4·1·(-118)
Δ = 481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{481}}{2*1}=\frac{-3-\sqrt{481}}{2}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{481}}{2*1}=\frac{-3+\sqrt{481}}{2}
$