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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2+5n+2=0

a = 2; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·2·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*2}=\frac{-8}{4}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*2}=\frac{-2}{4}
=-1/2
$