1. **State the problem:** Solve the inequality $$(\sqrt{x} + 2)^2 < (5 - \sqrt{x})^2$$. 2. **Recall the formula:** For any real numbers $a$ and $b$, $a^2 < b^2$ implies either $a < b$ and $a > -b$ or $a > b$ and $a < -b$. But since both sides involve square roots and are positive expressions, we can simplify by expanding and analyzing. 3. **Expand both sides:** $$ (\sqrt{x} + 2)^2 = (\sqrt{x})^2 + 2 \cdot 2 \cdot \sqrt{x} + 2^2 = x + 4\sqrt{x} + 4 $$ $$ (5 - \sqrt{x})^2 = 5^2 - 2 \cdot 5 \cdot \sqrt{x} + (\sqrt{x})^2 = 25 - 10\sqrt{x} + x $$ 4. **Rewrite the inequality:** $$ x + 4\sqrt{x} + 4 < 25 - 10\sqrt{x} + x $$ 5. **Subtract $x$ from both sides:** $$ \cancel{x} + 4\sqrt{x} + 4 < 25 - 10\sqrt{x} + \cancel{x} $$ $$ 4\sqrt{x} + 4 < 25 - 10\sqrt{x} $$ 6. **Bring all terms to one side:** $$ 4\sqrt{x} + 10\sqrt{x} + 4 - 25 < 0 $$ $$ 14\sqrt{x} - 21 < 0 $$ 7. **Add 21 to both sides:** $$ 14\sqrt{x} < 21 $$ 8. **Divide both sides by 14:** $$ \cancel{14}\sqrt{x} < \frac{21}{\cancel{14}} $$ $$ \sqrt{x} < \frac{21}{14} = \frac{3}{2} $$ 9. **Square both sides (both sides non-negative):** $$ x < \left(\frac{3}{2}\right)^2 = \frac{9}{4} $$ 10. **Domain restriction:** Since $\sqrt{x}$ is defined only for $x \geq 0$, the solution must satisfy: $$ 0 \leq x < \frac{9}{4} $$ **Final answer:** $$ \boxed{0 \leq x < \frac{9}{4}} $$