1. **Problem:** Solve $a^3+a^2=36$ for $a$. 2. **Formula used:** First, rewrite the equation in standard form and factor when possible. $$a^3+a^2-36=0$$ 3. **Important rule:** If an equation can be factored, set each factor equal to $0$. 4. Factor the expression: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 5. Set each factor equal to $0$: $$a-3=0$$ $$a^2+4a+12=0$$ 6. Solve the first factor: $$a=3$$ 7. Check the quadratic factor using the discriminant: $$b^2-4ac=4^2-4(1)(12)=16-48=-32$$ 8. Since the discriminant is negative, the quadratic has no real solution. 9. So the real solution is: $$a=3$$