First, expand the square of (x^2-3):

(x^2-3)^2 = x^4 - 6x^2 + 9

Now rewrite the equation with the expanded expression:
x^4 - 6x^2 + 9 + x^2 - 3 = 2

Combine like terms:
x^4 - 5x^2 + 6 = 2

Subtract 2 from both sides:
x^4 - 5x^2 + 4 = 0

Now, let y = x^2:
y^2 - 5y + 4 = 0

Factor the quadratic equation:
(y - 4)(y - 1) = 0
y = 4 or y = 1

Since y = x^2, solve for x:
x^2 = 4 or x^2 = 1

x = 2, x = -2, x = 1, or x = -1

Therefore, the solutions are x = 2, x = -2, x = 1, or x = -1.