2+2n+3n+4n+n*n=3000

Solution for 2+2n+3n+4n+n*n=3000 equation:

2+2n+3n+4n+n*n=3000

We move all terms to the left:

2+2n+3n+4n+n*n-(3000)=0

We add all the numbers together, and all the variables

9n+n*n-2998=0

Wy multiply elements

n^2+9n-2998=0

a = 1; b = 9; c = -2998;
Δ = b2-4ac
Δ = 92-4·1·(-2998)
Δ = 12073
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{-(9)-\sqrt{12073}}{2*1}=\frac{-9-\sqrt{12073}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{12073}}{2*1}=\frac{-9+\sqrt{12073}}{2}
$