1. **State the problem.** We need to solve the equation $a^3+a^2=36$ and find $a$. 2. **Rewrite the equation in standard form.** Move everything to one side: $$a^3+a^2-36=0$$ 3. **Look for an integer root.** Since $36$ is a small number, test simple integer values. Try $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ is a solution. 4. **Factor using the known root.** Because $a=3$ works, $(a-3)$ is a factor: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ So the equation becomes: $$ (a-3)(a^2+4a+12)=0 $$ 5. **Set each factor equal to zero.** From $(a-3)=0$: $$a=3$$ 6. **Check the quadratic factor.** Solve $a^2+4a+12=0$ using the discriminant: $$b^2-4ac=4^2-4(1)(12)=16-48=-32$$ Since the discriminant is negative, this quadratic has no real solutions. 7. **Final answer.** The real solution is: $$a=3$$