(2+2/5)n=-4/15

Solution for (2+2/5)n=-4/15 equation:

(2+2/5)n=-4/15

We move all terms to the left:

(2+2/5)n-(-4/15)=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

(2/5+2)n-(-4/15)=0

We multiply parentheses

2n^2+2n-(-4/15)=0

We get rid of parentheses

2n^2+2n+4/15=0

We multiply all the terms by the denominator


2n^2*15+2n*15+4=0

Wy multiply elements

30n^2+30n+4=0

a = 30; b = 30; c = +4;
Δ = b2-4ac
Δ = 302-4·30·4
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{105}}{2*30}=\frac{-30-2\sqrt{105}}{60}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{105}}{2*30}=\frac{-30+2\sqrt{105}}{60}
$