N^3-3n^2+2n=336

Solution for N^3-3n^2+2n=336 equation:

^3-3N^2+2N=336

We move all terms to the left:

^3-3N^2+2N-(336)=0

determiningTheFunctionDomain
-3N^2+2N-336+^3=0

We add all the numbers together, and all the variables

-3N^2+2N=0

a = -3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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