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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

2(3n-1)n

We multiply parentheses

6n^2-2n

Back to the equation:

-(6n^2-2n)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6n^2+8n-2=0

a = -6; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·(-6)·(-2)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-6}=\frac{-12}{-12}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-6}=\frac{-4}{-12}
=1/3
$