Solve for n (2n)!=30(2n-2)! n is in N set

Here is the given
equation:

(2n)! =
30(2n-2)!

Since (2n)! =
(2n)(2n-1)(2n-2)!,

we can plug this into the above
equation.

=> (2n)(2n-1)(2n-2)! =
30(2n-2)!

When we divide each side by
(2n-2)!,

we get (2n)(2n-1) =
30,

=> 4n^2 - 2n =
30

=> 4n^2 - 2n - 30 =
0

=> (4n+10)(n-3) =
0

Therefore, n= -5/2 or
3

However, since only n=3 is in the N
set,

n=3 will be the only
answer.