Let's make a substitution to simplify the equation:

Let y = 2^(x^2+2), then the equation becomes:

4y^2 - 9y + 8 = 0

Now, we can factor this quadratic equation:

(4y - 8)(y - 1) = 0

Solving for y by setting each factor to zero:

4y - 8 = 0 or y - 1 = 0
4y = 8 or y = 1
y = 2 or y = 1

Now, substitute back y = 2^(x^2+2) and y = 1:

2^(x^2+2) = 2 or 2^(x^2+2) = 1

For 2^(x^2+2) = 2, we have:
x^2 + 2 = 1
x^2 = -1
x = ±i

For 2^(x^2+2) = 1, we have:
x^2 + 2 = 0
x^2 = -2
x = ±√(-2)
x = ±i√2

Therefore, the solutions to the equation are:
x = i, x = -i, x = i√2, x = -i√2.