(5/6)-(2/3)n=(3/2)

Solution for (5/6)-(2/3)n=(3/2) equation:

(5/6)-(2/3)n=(3/2)

We move all terms to the left:

(5/6)-(2/3)n-((3/2))=0


Domain of the equation: 3)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

-(+2/3)n+(+5/6)-((+3/2))=0

We multiply parentheses

-2n^2+(+5/6)-((+3/2))=0

We get rid of parentheses

-2n^2+5/6-((+3/2))=0

We calculate fractions


-2n^2+()/()+()/()=0

We add all the numbers together, and all the variables

-2n^2+2=0

a = -2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-2)·2
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-2}=\frac{-4}{-4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-2}=\frac{4}{-4}
=-1
$