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.

By definition, (2n)! = (2n-2)!*(2n-1)*2n

We'll re-write the equation:

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

We'll divide by (2n-2)!:

(2n-1)*2n = 30

We'll divide by 2:

n(2n - 1) = 15

We'll remove the brackets:

2n^2 - n - 15 = 0

We'll apply quadratic formula:

n1 = [1 + sqrt(1 + 120)]/4

n1 = (1 + 11)/4

n1 = 3

n2 = -10/4

n2 = -5/2

**Since just n1= 3 is an element of N set, we'll reject the value of n2. The equation has only one solution, n = 3.**