2n(n-1)+4n=2(3n-1)

Solution for 2n(n-1)+4n=2(3n-1) equation:

2n(n-1)+4n=2(3n-1)

We move all terms to the left:

2n(n-1)+4n-(2(3n-1))=0

We add all the numbers together, and all the variables

4n+2n(n-1)-(2(3n-1))=0

We multiply parentheses

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


We calculate terms in parentheses: -(2(3n-1)), so:

2(3n-1)

We multiply parentheses

6n-2

Back to the equation:

-(6n-2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

2n^2+2n-6n+2=0

We add all the numbers together, and all the variables

2n^2-4n+2=0

a = 2; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·2·2
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$n=\frac{-b}{2a}=\frac{4}{4}=1$