N^3-3n^2+2n=(n-8)

Solution for N^3-3n^2+2n=(n-8) equation:

^3-3N^2+2N=(N-8)

We move all terms to the left:

^3-3N^2+2N-((N-8))=0

We add all the numbers together, and all the variables

-3N^2+2N-((N-8))=0


We calculate terms in parentheses: -((N-8)), so:

(N-8)

We get rid of parentheses

N-8

Back to the equation:

-(N-8)


We get rid of parentheses

-3N^2+2N-N+8=0

We add all the numbers together, and all the variables

-3N^2+N+8=0

a = -3; b = 1; c = +8;
Δ = b2-4ac
Δ = 12-4·(-3)·8
Δ = 97
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{-(1)-\sqrt{97}}{2*-3}=\frac{-1-\sqrt{97}}{-6}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{97}}{2*-3}=\frac{-1+\sqrt{97}}{-6}
$