(1+n)n=999

Solution for (1+n)n=999 equation:

(1+n)n=999

We move all terms to the left:

(1+n)n-(999)=0

We add all the numbers together, and all the variables

(n+1)n-999=0

We multiply parentheses

n^2+n-999=0

a = 1; b = 1; c = -999;
Δ = b2-4ac
Δ = 12-4·1·(-999)
Δ = 3997
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{-(1)-\sqrt{3997}}{2*1}=\frac{-1-\sqrt{3997}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3997}}{2*1}=\frac{-1+\sqrt{3997}}{2}
$