1. **State the problem:** Solve the equation $y^2 + y^3 = 12$ for $y$. 2. **Rewrite the equation:** The equation can be written as $$y^3 + y^2 = 12$$ 3. **Factor the left side:** Factor out the common term $y^2$: $$y^2(y + 1) = 12$$ 4. **Isolate terms:** We want to find values of $y$ such that this holds true. 5. **Try possible rational roots:** Since the equation is cubic, try simple integer values like $y=1, 2, 3$: - For $y=1$: $1^3 + 1^2 = 1 + 1 = 2 \neq 12$ - For $y=2$: $2^3 + 2^2 = 8 + 4 = 12$ which satisfies the equation. 6. **Check for other roots:** Divide the cubic polynomial $y^3 + y^2 - 12 = 0$ by $(y-2)$: $$\frac{y^3 + y^2 - 12}{y - 2} = y^2 + 3y + 6$$ 7. **Solve the quadratic:** Set $$y^2 + 3y + 6 = 0$$ Use the quadratic formula: $$y = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times 6}}{2 \times 1} = \frac{-3 \pm \sqrt{9 - 24}}{2} = \frac{-3 \pm \sqrt{-15}}{2}$$ 8. **Interpret roots:** The discriminant is negative, so the quadratic has no real roots. 9. **Final answer:** The only real solution is $$y = 2$$