1/3n+4-1=n-1

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

1/3n+4-1=n-1

We move all terms to the left:

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


Domain of the equation: 3n!=0

n!=0/3

n!=0

n∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/3n-n+1+3=0

We multiply all the terms by the denominator


-n*3n+1*3n+3*3n+1=0

Wy multiply elements

-3n^2+3n+9n+1=0

We add all the numbers together, and all the variables

-3n^2+12n+1=0

a = -3; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-3)·1
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{39}}{2*-3}=\frac{-12-2\sqrt{39}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{39}}{2*-3}=\frac{-12+2\sqrt{39}}{-6}
$