1. **Stating the problem:** Solve the inequality $$2^{2x+2} < 6^x + 2 \cdot 3^{2x+2}$$. 2. **Rewrite terms with common bases:** Note that $$6^x = (2 \cdot 3)^x = 2^x \cdot 3^x$$ and $$3^{2x+2} = 3^2 \cdot 3^{2x} = 9 \cdot (3^2)^x = 9 \cdot 3^{2x}$$. 3. **Express all terms in terms of powers of 2 and 3:** $$2^{2x+2} = 2^2 \cdot 2^{2x} = 4 \cdot (2^2)^x = 4 \cdot 4^x$$ 4. **Rewrite inequality:** $$4 \cdot 4^x < 2^x \cdot 3^x + 2 \cdot 9 \cdot 3^{2x}$$ which simplifies to $$4 \cdot 4^x < 2^x 3^x + 18 \cdot 3^{2x}$$. 5. **Substitute:** Let $$a = 2^x$$ and $$b = 3^x$$, so the inequality becomes $$4 \cdot a^2 < a b + 18 b^2$$. 6. **Rewrite inequality:** $$4 a^2 - a b - 18 b^2 < 0$$. 7. **Divide both sides by $$b^2$$ (positive since $$b=3^x > 0$$):** $$4 \frac{a^2}{b^2} - \frac{a}{b} - 18 < 0$$ Intermediate step with cancellation: $$4 \cancel{\frac{a^2}{b^2}} - \cancel{\frac{a}{b}} - 18 < 0$$ (just showing division by $$b^2$$) 8. **Let $$t = \frac{a}{b} = \frac{2^x}{3^x} = \left(\frac{2}{3}\right)^x$$, so the inequality becomes $$4 t^2 - t - 18 < 0$$. 9. **Solve quadratic inequality:** The quadratic is $$4 t^2 - t - 18 = 0$$. Calculate discriminant: $$\Delta = (-1)^2 - 4 \cdot 4 \cdot (-18) = 1 + 288 = 289$$. Roots: $$t = \frac{1 \pm \sqrt{289}}{2 \cdot 4} = \frac{1 \pm 17}{8}$$. So, $$t_1 = \frac{1 - 17}{8} = -2$$ (discard negative root since $$t > 0$$), $$t_2 = \frac{1 + 17}{8} = \frac{18}{8} = 2.25$$. 10. **Since $$4 t^2 - t - 18$$ opens upward, inequality $$< 0$$ holds between roots, so $$0 < t < 2.25$$. 11. **Recall $$t = \left(\frac{2}{3}\right)^x$$, so $$0 < \left(\frac{2}{3}\right)^x < 2.25$$. Since $$\left(\frac{2}{3}\right)^x > 0$$ for all real $$x$$, the left inequality is always true. 12. **Solve $$\left(\frac{2}{3}\right)^x < 2.25$$:** Take natural logarithm: $$x \ln\left(\frac{2}{3}\right) < \ln(2.25)$$. Note $$\ln\left(\frac{2}{3}\right) < 0$$, so inequality reverses: $$x > \frac{\ln(2.25)}{\ln(\frac{2}{3})}$$. Calculate approximate values: $$\ln(2.25) \approx 0.81093$$, $$\ln(\frac{2}{3}) \approx -0.40547$$, so $$x > \frac{0.81093}{-0.40547} = -2$$. 13. **Final solution:** $$\boxed{x > -2}$$. This means all real $$x$$ greater than $$-2$$ satisfy the original inequality.