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

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

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

We move all terms to the left:

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

We multiply parentheses

8n^2-8n-(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 get rid of parentheses

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

We add all the numbers together, and all the variables

8n^2-14n+2=0

a = 8; b = -14; c = +2;
Δ = b2-4ac
Δ = -142-4·8·2
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{33}}{2*8}=\frac{14-2\sqrt{33}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{33}}{2*8}=\frac{14+2\sqrt{33}}{16}
$