1. **State the problem:** Solve the inequality $$4x^2 + 12x + 9 > 0$$. 2. **Identify the type of inequality:** This is a quadratic inequality. We will analyze the quadratic expression and determine where it is greater than zero. 3. **Recall the quadratic formula:** For a quadratic equation $$ax^2 + bx + c = 0$$, the roots are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 12^2 - 4 \times 4 \times 9 = 144 - 144 = 0$$. 5. **Interpret the discriminant:** Since $$\Delta = 0$$, the quadratic has one repeated root. 6. **Find the root:** $$x = \frac{-12}{2 \times 4} = \frac{-12}{8} = -\frac{3}{2}$$. 7. **Rewrite the quadratic:** Since the discriminant is zero, the quadratic can be written as a perfect square: $$4x^2 + 12x + 9 = (2x + 3)^2$$. 8. **Analyze the inequality:** We want to find where $$(2x + 3)^2 > 0$$. 9. **Important rule:** A square of a real number is always $$\geq 0$$, and equals zero only when the base is zero. 10. **Determine solution:** The expression is zero at $$x = -\frac{3}{2}$$ and positive everywhere else. 11. **Final solution:** $$x \in (-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, \infty)$$. In other words, the inequality holds for all real numbers except $$x = -\frac{3}{2}$$.