(n)=-6-+(n-1)(1/5)

Solution for (n)=-6-+(n-1)(1/5) equation:

(n)=-6-+(n-1)(1/5)

We move all terms to the left:

(n)-(-6-+(n-1)(1/5))=0

We add all the numbers together, and all the variables

n-(-6-+(n-1)(+1/5))=0

We use the square of the difference formula

n-(-6-(n-1)(+1/5))=0

We multiply parentheses ..

-(-6-(+n^2-1*1/5))+n=0

We multiply all the terms by the denominator


-(-6-(+n^2-1*1+n*5))=0


We calculate terms in parentheses: -(-6-(+n^2-1*1+n*5)), so:

-6-(+n^2-1*1+n*5)

determiningTheFunctionDomain
-(+n^2-1*1+n*5)-6

We get rid of parentheses

-n^2-n*5-6+1*1

We add all the numbers together, and all the variables

-1n^2-n*5-5

Wy multiply elements

-1n^2-5n-5

Back to the equation:

-(-1n^2-5n-5)


We get rid of parentheses

1n^2+5n+5=0

We add all the numbers together, and all the variables

n^2+5n+5=0

a = 1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·1·5
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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