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

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

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

We move all terms to the left:

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

We multiply parentheses

8n^2-8n-((3n-1))=0


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

(3n-1)

We get rid of parentheses

3n-1

Back to the equation:

-(3n-1)


We get rid of parentheses

8n^2-8n-3n+1=0

We add all the numbers together, and all the variables

8n^2-11n+1=0

a = 8; b = -11; c = +1;
Δ = b2-4ac
Δ = -112-4·8·1
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{89}}{2*8}=\frac{11-\sqrt{89}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{89}}{2*8}=\frac{11+\sqrt{89}}{16}
$