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

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

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: +((-1N+2)+6N-3N^2), so:

(-1N+2)+6N-3N^2

determiningTheFunctionDomain
-3N^2+(-1N+2)+6N

We add all the numbers together, and all the variables

-3N^2+6N+(-1N+2)

We get rid of parentheses

-3N^2+6N-1N+2

We add all the numbers together, and all the variables

-3N^2+5N+2

Back to the equation:

+(-3N^2+5N+2)


We get rid of parentheses

-3N^2+5N+2=0

a = -3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·(-3)·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-3}=\frac{-12}{-6}
=+2
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-3}=\frac{2}{-6}
=-1/3
$