(n/3)+(3/4)=(5/6)n-1

Solution for (n/3)+(3/4)=(5/6)n-1 equation:

(n/3)+(3/4)=(5/6)n-1

We move all terms to the left:

(n/3)+(3/4)-((5/6)n-1)=0


Domain of the equation: 6)n-1)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+n/3)-((+5/6)n-1)+(+3/4)=0

We get rid of parentheses

n/3-((+5/6)n-1)+3/4=0

We calculate fractions


24n^2/72n+()/72n+324n/72n=0

We multiply all the terms by the denominator


24n^2+324n+()=0

We add all the numbers together, and all the variables

24n^2+324n=0

a = 24; b = 324; c = 0;
Δ = b2-4ac
Δ = 3242-4·24·0
Δ = 104976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{104976}=324$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(324)-324}{2*24}=\frac{-648}{48}
=-13+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(324)+324}{2*24}=\frac{0}{48}
=0
$