n+2/3n=2n-4

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

n+2/3n=2n-4

We move all terms to the left:

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


Domain of the equation: 3n!=0

n!=0/3

n!=0

n∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

3n^2-6n^2+12n+2=0

We add all the numbers together, and all the variables

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

a = -3; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·(-3)·2
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*-3}=\frac{-12-2\sqrt{42}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*-3}=\frac{-12+2\sqrt{42}}{-6}
$