1. **State the problem:** Solve the quadratic equation $$(3x + 3)^2 + 6(3x + 3) + 5 = 0.$$\n\n2. **Use substitution:** Let $y = 3x + 3$. The equation becomes $$y^2 + 6y + 5 = 0.$$\n\n3. **Solve the quadratic in $y$:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=6$, and $c=5$.\n\n4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 6^2 - 4 \times 1 \times 5 = 36 - 20 = 16.$$\n\n5. **Find the roots for $y$:** $$y = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2}.$$\n\n6. **Calculate each root:**\n- For the plus sign: $$y = \frac{-6 + 4}{2} = \frac{-2}{2} = -1.$$\n- For the minus sign: $$y = \frac{-6 - 4}{2} = \frac{-10}{2} = -5.$$\n\n7. **Back-substitute $y = 3x + 3$:**\n- When $y = -1$: $$3x + 3 = -1 \Rightarrow 3x = -1 - 3 = -4 \Rightarrow x = \frac{\cancel{3}x}{\cancel{3}} = \frac{-4}{3}.$$\n- When $y = -5$: $$3x + 3 = -5 \Rightarrow 3x = -5 - 3 = -8 \Rightarrow x = \frac{\cancel{3}x}{\cancel{3}} = \frac{-8}{3}.$$\n\n8. **Final answer:** The solutions are $$x = -\frac{4}{3}$$ and $$x = -\frac{8}{3}.$$