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

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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} $

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