Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

Let t = x^2 - 4.

Let a = √3 + √2. Then √3 - √2 = 1/a (since (√3+√2)(√3-√2) = 3 - 2 = 1).

So the equation becomes:

a^t - a^(-t) = 10.

Let y = a^t > 0.

Then y - 1/y = 10

=> y^2 - 10y - 1 = 0.

Solving, y = 5 ± √26.

But only y = 5 + √26 > 0 is valid.

So a^t = 5 + √26

=> t = ln(5 + √26) / ln(a).

Now, x^2 = t + 4.

Thus there are two real values of x:

x = ±√(t + 4).

Therefore, n(S) = 2.