1. The problem is to solve the equation $x^2 = 9$ over the real numbers $\mathbb{R}$. 2. The general formula to solve $x^2 = a$ is: $$x = \pm \sqrt{a}$$ This means $x$ can be either the positive or negative square root of $a$, provided $a \geq 0$. 3. Important rule: If $a < 0$, there are no real solutions because the square of a real number cannot be negative. 4. Applying this to $x^2 = 9$: $$x = \pm \sqrt{9}$$ 5. Calculate the square root: $$x = \pm 3$$ 6. Therefore, the solutions are: $$x = 3 \quad \text{or} \quad x = -3$$ Final answer: $x = \pm 3$