1. **State the problem:** Solve for $a$ in the equation $$a^3 + a^2 = 36.$$\n\n2. **Rewrite the equation:** Factor out the common term $a^2$ from the left side:\n$$a^3 + a^2 = a^2(a + 1) = 36.$$\n\n3. **Analyze the factored form:** We have $$a^2(a + 1) = 36.$$ We want to find values of $a$ that satisfy this.\n\n4. **Try possible integer values:** Since $36$ is a positive integer, try integer values for $a$ to see if the equation holds.\n- For $a=2$: $$2^2(2+1) = 4 \times 3 = 12 \neq 36.$$\n- For $a=3$: $$3^2(3+1) = 9 \times 4 = 36.$$ This satisfies the equation.\n\n5. **Check for other solutions:**\nRewrite as $$a^2(a+1) - 36 = 0.$$\nThis is a cubic equation: $$a^3 + a^2 - 36 = 0.$$\n\n6. **Use Rational Root Theorem:** Possible rational roots are factors of 36: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$. We already found $a=3$ works.\n\n7. **Divide polynomial by $(a-3)$:**\n$$\frac{a^3 + a^2 - 36}{a-3} = a^2 + 4a + 12.$$\n\n8. **Solve quadratic $a^2 + 4a + 12 = 0$:**\nUse quadratic formula: $$a = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 12}}{2} = \frac{-4 \pm \sqrt{16 - 48}}{2} = \frac{-4 \pm \sqrt{-32}}{2}.$$\n\n9. **Simplify the discriminant:**\n$$\sqrt{-32} = \sqrt{32}i = 4\sqrt{2}i.$$\n\n10. **Final complex roots:**\n$$a = \frac{-4 \pm 4\sqrt{2}i}{2} = -2 \pm 2\sqrt{2}i.$$\n\n**Answer:** The real solution is $$\boxed{3}$$ and the two complex solutions are $$-2 + 2\sqrt{2}i$$ and $$-2 - 2\sqrt{2}i.$$