(n(2n+n))/2=O(n^2)

Solution for (n(2n+n))/2=O(n^2) equation:

(n(2n+n))/2=(n^2)

We move all terms to the left:

(n(2n+n))/2-((n^2))=0

determiningTheFunctionDomain
(n(2n+n))/2-n^2=0

We add all the numbers together, and all the variables

-n^2+(n(+3n))/2=0

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


-1n^2*2+(n(+3n))=0


We calculate terms in parentheses: +(n(+3n)), so:

n(+3n)

We multiply parentheses

3n^2

Back to the equation:

+(3n^2)


We add all the numbers together, and all the variables

3n^2-1n^2*2=0

Wy multiply elements

3n^2-2n^2=0

We add all the numbers together, and all the variables

n^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$n=\frac{-b}{2a}=\frac{0}{2}=0$