19/6+13/4n=2/3n+1

Solution for 19/6+13/4n=2/3n+1 equation:

19/6+13/4n=2/3n+1

We move all terms to the left:

19/6+13/4n-(2/3n+1)=0


Domain of the equation: 4n!=0

n!=0/4

n!=0

n∈R



Domain of the equation: 3n+1)!=0

n∈R


We get rid of parentheses

13/4n-2/3n-1+19/6=0

We calculate fractions


684n^2/432n^2+1404n/432n^2+(-288n)/432n^2-1=0

We multiply all the terms by the denominator


684n^2+1404n+(-288n)-1*432n^2=0

Wy multiply elements

684n^2-432n^2+1404n+(-288n)=0

We get rid of parentheses

684n^2-432n^2+1404n-288n=0

We add all the numbers together, and all the variables

252n^2+1116n=0

a = 252; b = 1116; c = 0;
Δ = b2-4ac
Δ = 11162-4·252·0
Δ = 1245456
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1245456}=1116$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1116)-1116}{2*252}=\frac{-2232}{504}
=-4+3/7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1116)+1116}{2*252}=\frac{0}{504}
=0
$