(2/3)n-(1/3)n+6=11

Solution for (2/3)n-(1/3)n+6=11 equation:

(2/3)n-(1/3)n+6=11

We move all terms to the left:

(2/3)n-(1/3)n+6-(11)=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-(+1/3)n+6-11=0

We add all the numbers together, and all the variables

(+2/3)n-(+1/3)n-5=0

We multiply parentheses

2n^2-n^2-5=0

We add all the numbers together, and all the variables

n^2-5=0

a = 1; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·1·(-5)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*1}=\frac{0-2\sqrt{5}}{2}
=-\frac{2\sqrt{5}}{2}
=-\sqrt{5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*1}=\frac{0+2\sqrt{5}}{2}
=\frac{2\sqrt{5}}{2}
=\sqrt{5}
$