1. State the problem: Solve for $a$ in $a^3+a^2=36$. 2. Set up the equation. $$a^3+a^2=36$$ 3. Factor the left side. $$a^3+a^2=a^2(a+1)$$ So, $$a^2(a+1)=36$$ 4. Move everything to one side. $$a^2(a+1)-36=0$$ $$a^3+a^2-36=0$$ 5. Find rational roots by testing factors of $36$. Check $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ is a solution. 6. Factor using the found root $a=3$. Divide the polynomial $a^3+a^2-36$ by $(a-3)$: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 7. Solve the remaining factor. $$a^2+4a+12=0$$ Use the quadratic formula $a=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=1,b=4,c=12$. $$a=\frac{-4\pm\sqrt{4^2-4(1)(12)}}{2(1)}$$ $$a=\frac{-4\pm\sqrt{16-48}}{2}$$ $$a=\frac{-4\pm\sqrt{-32}}{2}$$ 8. Simplify the square root of a negative number. $$a=\frac{-4\pm 4\sqrt{2}i}{2}$$ $$a=-2\pm 2\sqrt{2}i$$ 9. Final answer (real solutions only if you want real $a$): $\boxed{a=3}$ for real $a$. 10. If complex solutions are allowed, then: $\boxed{a=3,\;a=-2\pm 2\sqrt{2}i}$.