(3/5)n+2=17

Solution for (3/5)n+2=17 equation:

(3/5)n+2=17

We move all terms to the left:

(3/5)n+2-(17)=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

(+3/5)n+2-17=0

We add all the numbers together, and all the variables

(+3/5)n-15=0

We multiply parentheses

3n^2-15=0

a = 3; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·3·(-15)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*3}=\frac{0-6\sqrt{5}}{6}
=-\frac{6\sqrt{5}}{6}
=-\sqrt{5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*3}=\frac{0+6\sqrt{5}}{6}
=\frac{6\sqrt{5}}{6}
=\sqrt{5}
$