1. **Stating the problem:** Find the smallest real solutions of the equation $$x^2 = g$$ where $$g$$ is a real number. 2. **Formula used:** The solutions to $$x^2 = g$$ are given by $$x = \pm \sqrt{g}$$, provided that $$g \geq 0$$ because the square root of a negative number is not a real number. 3. **Important rule:** For real solutions, $$g$$ must be non-negative. If $$g < 0$$, there are no real solutions. 4. **Finding the smallest real solution:** The two solutions are $$x = +\sqrt{g}$$ and $$x = -\sqrt{g}$$. Since $$-\sqrt{g} < +\sqrt{g}$$, the smallest real solution is $$x = -\sqrt{g}$$. 5. **Summary:** The smallest real solution to $$x^2 = g$$ is $$x = -\sqrt{g}$$, assuming $$g \geq 0$$. **Final answer:** $$\boxed{x = -\sqrt{g}}$$