You will find critical points where f'(x) = 0. So:

f'(x) = 5x^4 - 15x^2 - 20 = 0

This is a "bi-quadratic" equation. To make things look simpler, let y = x^2. Then we need to solve:

5y^2 - 15y - 20 = 0

5(y^2 - 3y - 4) = 0

5(y - 4)(y + 1) = 0

So y = 4 and y = -1.

Now, y = x^2 so we have x^2 = 4 and x^2 = -1.

x^2 = 4

x = -2 and x = 2.

x^2 = -1

has no real solutions.

So the critical points are at x = -2, 2.

-Dan