1. State the problem: Solve for $a$ in $a^3+a^2=36$.\n 2. Write the equation and factor it: $$a^3+a^2=36$$\n 3. Factor out $a^2$: $$a^2(a+1)=36$$\n 4. Move everything to one side: $$a^3+a^2-36=0$$\n 5. Try a simple integer solution. Test $a=3$: $$3^3+3^2=27+9=36$$\n 6. Since $a=3$ works, factor $(a-3)$ out. Divide $a^3+a^2-36$ by $(a-3)$ to get: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$\n 7. Solve the remaining quadratic: $$a^2+4a+12=0$$\n 8. Use the quadratic formula: $$a=\frac{-4\pm\sqrt{4^2-4\cdot 1\cdot 12}}{2\cdot 1}$$\n 9. Simplify the discriminant: $$4^2-4\cdot 1\cdot 12=16-48=-32$$\n 10. Continue simplifying (no cancellation here): $$a=\frac{-4\pm\sqrt{-32}}{2}$$\n 11. Write $ \sqrt{-32}=\sqrt{32}\,i=4\sqrt{2}\,i$: $$a=\frac{-4\pm 4\sqrt{2}\,i}{2}$$\n 12. Divide numerator and denominator by $2$ (show cancellation): $$a=\cancel{\frac{2}{2}}\left(-2\pm 2\sqrt{2}\,i\right)=-2\pm 2\sqrt{2}\,i$$\n 13. Final answer: $$a=3\ \text{or}\ a=-2\pm 2\sqrt{2}\,i$$\n