(n+2)+n(n+3)+n=0

Solution for (n+2)+n(n+3)+n=0 equation:

(n+2)+n(n+3)+n=0

We add all the numbers together, and all the variables

n+(n+2)+n(n+3)=0

We multiply parentheses

n^2+n+(n+2)+3n=0

We get rid of parentheses

n^2+n+n+3n+2=0

We add all the numbers together, and all the variables

n^2+5n+2=0

a = 1; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·1·2
Δ = 17
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{-(5)-\sqrt{17}}{2*1}=\frac{-5-\sqrt{17}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{17}}{2*1}=\frac{-5+\sqrt{17}}{2}
$