N2! ?? , so that would be like 2N ??
let N2! = x
3(x+24) = 2x
3x + 72 = 2x
x = -72
N2! = -72
N = -36
If 3(N2! +24)=2N2!, find the positive value of N?
2 answers
If you mean
3((n^2)! + 24) = 2(n^2)!
then let u = (n^2)! and you have
3u + 72 = 2u
which has no solutions, since (n^2)! is never negative.
Care to repost using some parentheses to clarify what you mean?
3((n^2)! + 24) = 2(n^2)!
then let u = (n^2)! and you have
3u + 72 = 2u
which has no solutions, since (n^2)! is never negative.
Care to repost using some parentheses to clarify what you mean?
1 answer