(N^3+6n^2-4n+1)/(n+1)=0

Solution for (N^3+6n^2-4n+1)/(n+1)=0 equation:

(^3+6N^2-4N+1)/(N+1)=0


Domain of the equation: (N+1)!=0

We move all terms containing N to the left, all other terms to the right

N!=-1

N∈R


We multiply all the terms by the denominator


(^3+6N^2-4N+1)=0

We get rid of parentheses

6N^2-4N+1+^3=0

We add all the numbers together, and all the variables

6N^2-4N=0

a = 6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·6·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*6}=\frac{0}{12}
=0
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*6}=\frac{8}{12}
=2/3
$