Choice D is the correct answer.

1. Subtract 9 from both sides: $x^2 = -9$.
2. Take the square root: $x = \pm \sqrt{-9}$.

Since the square of any real number is non-negative, there is no real number that squares to -9. The solutions are imaginary ($\pm 3i$).

Choice A is incorrect because $3^2 = 9 eq -9$. Choice B is incorrect because $(-3)^2 = 9 eq -9$. Choice C is incorrect because $\pm 3$ are solutions to $x^2 - 9 = 0$.