1. State the problem: Solve for $a$ in $a^3+a^2=36$.\n 2. Use the given equation: $a^3+a^2=36$.\n 3. Factor the left side: $$a^3+a^2=a^2(a+1).$$\n 4. Set up the factored equation: $a^2(a+1)=36$.\n 5. Move to one side and write as a polynomial equation: $$a^3+a^2-36=0.$$\n 6. Try rational roots (possible values from factors of $36$ over factors of $1$): $\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$.\n 7. Test $a=3$: $$3^3+3^2=27+9=36,$$ so $a=3$ is a solution.\n 8. Factor using the root $a=3$: $$a^3+a^2-36=(a-3)(a^2+4a+12).$$\n 9. Use the rule for a product to be zero: $$ (a-3)(a^2+4a+12)=0. $$\n 10. Solve each factor. First, $a-3=0 \Rightarrow a=3$.\n 11. For the quadratic, set $a^2+4a+12=0$.\n 12. Use the quadratic formula: $$a=\frac{-4\pm\sqrt{4^2-4(1)(12)}}{2(1)}.$$\n 13. Simplify the discriminant: $$4^2-4(1)(12)=16-48=-32.$$\n 14. Conclude there are no real roots because the discriminant is negative: $$a=\frac{-4\pm\sqrt{-32}}{2}=-2\pm i\sqrt{8}=-2\pm 2\sqrt{2}\,i.$$\n 15. Final answer (real solutions): $a=3$.