66=n(n-3)/2+n

Solution for 66=n(n-3)/2+n equation:

66=n(n-3)/2+n

We move all terms to the left:

66-(n(n-3)/2+n)=0


Domain of the equation: 2+n)!=0

We move all terms containing n to the left, all other terms to the right

n)!=-2

n!=-2/1

n!=-2

n∈R


We multiply all the terms by the denominator


-(n(n-3)+66*2+n)=0


We calculate terms in parentheses: -(n(n-3)+66*2+n), so:

n(n-3)+66*2+n

determiningTheFunctionDomain
n(n-3)+n+66*2

We add all the numbers together, and all the variables

n+n(n-3)+132

We multiply parentheses

n^2+n-3n+132

We add all the numbers together, and all the variables

n^2-2n+132

Back to the equation:

-(n^2-2n+132)


We get rid of parentheses

-n^2+2n-132=0

We add all the numbers together, and all the variables

-1n^2+2n-132=0

a = -1; b = 2; c = -132;
Δ = b2-4ac
Δ = 22-4·(-1)·(-132)
Δ = -524
Delta is less than zero, so there is no solution for the equation