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

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

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

We move all terms to the left:

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


Domain of the equation: 4n!=0

n!=0/4

n!=0

n∈R



Domain of the equation: 2n-1)!=0

n∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

3/4n+3n-2n-3-3=0

We multiply all the terms by the denominator


3n*4n-2n*4n-3*4n-3*4n+3=0

Wy multiply elements

12n^2-8n^2-12n-12n+3=0

We add all the numbers together, and all the variables

4n^2-24n+3=0

a = 4; b = -24; c = +3;
Δ = b2-4ac
Δ = -242-4·4·3
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{33}}{2*4}=\frac{24-4\sqrt{33}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{33}}{2*4}=\frac{24+4\sqrt{33}}{8}
$