1. State the problem: Solve for $a$ in the equation $a^3 + a^2 = 36$. 2. Factor the left side (common factor $a^2$): $$a^3+a^2=a^2(a+1)$$ 3. Rewrite the equation using the factorization: $$a^2(a+1)=36$$ 4. Bring everything to one side to solve a polynomial equation: $$a^2(a+1)-36=0$$ $$a^3+a^2-36=0$$ 5. Try simple integer values. Test $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ is a solution. 6. Factor using $(a-3)$ (since $a=3$ is a root). Divide $a^3+a^2-36$ by $(a-3)$ to get: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 7. Solve the remaining quadratic: $$a^2+4a+12=0$$ 8. Use the quadratic formula: $$a=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot12}}{2}$$ $$a=\frac{-4\pm\sqrt{16-48}}{2}$$ $$a=\frac{-4\pm\sqrt{-32}}{2}$$ 9. Simplify the square root: $$\sqrt{-32}=\sqrt{-16\cdot2}=4\sqrt{-2}=4i\sqrt{2}$$ $$a=\frac{-4\pm 4i\sqrt{2}}{2}$$ $$a=-2\pm 2i\sqrt{2}$$ 10. Final answer: $$a=3\quad\text{or}\quad a=-2\pm 2i\sqrt{2}$$