(1/5)n-3+(3/5)n=9

Solution for (1/5)n-3+(3/5)n=9 equation:

(1/5)n-3+(3/5)n=9

We move all terms to the left:

(1/5)n-3+(3/5)n-(9)=0


Domain of the equation: 5)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+1/5)n+(+3/5)n-3-9=0

We add all the numbers together, and all the variables

(+1/5)n+(+3/5)n-12=0

We multiply parentheses

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

We add all the numbers together, and all the variables

4n^2-12=0

a = 4; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4·(-12)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*4}=\frac{0-8\sqrt{3}}{8}
=-\frac{8\sqrt{3}}{8}
=-\sqrt{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*4}=\frac{0+8\sqrt{3}}{8}
=\frac{8\sqrt{3}}{8}
=\sqrt{3}
$