(-1/3)n+(-2)=-18

Solution for (-1/3)n+(-2)=-18 equation:

(-1/3)n+(-2)=-18

We move all terms to the left:

(-1/3)n+(-2)-(-18)=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

(-1/3)n+16=0

We multiply parentheses

-1n^2+16=0

a = -1; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-1)·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-1}=\frac{-8}{-2}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-1}=\frac{8}{-2}
=-4
$