1. **State the problem:** Solve for $a$ in the equation $$a^3 + a^2 = 36.$$\n\n2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$a^3 + a^2 - 36 = 0.$$\n\n3. **Factor the equation if possible:** First, check for rational roots using the Rational Root Theorem. Possible roots are factors of 36: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$.\n\n4. **Test $a=2$:** $$2^3 + 2^2 = 8 + 4 = 12 \neq 36.$$\n\n5. **Test $a=3$:** $$3^3 + 3^2 = 27 + 9 = 36.$$ This satisfies the equation, so $a=3$ is a root.\n\n6. **Divide the polynomial by $(a-3)$ to find other roots:**\nUse polynomial division or synthetic division: $$\frac{a^3 + a^2 - 36}{a - 3}.$$\n\n7. **Perform synthetic division:**\nCoefficients: 1 (for $a^3$), 1 (for $a^2$), 0 (for $a$), -36 (constant).\nMultiply and add steps: \n- Bring down 1.\n- Multiply 1 by 3 = 3, add to 1 = 4.\n- Multiply 4 by 3 = 12, add to 0 = 12.\n- Multiply 12 by 3 = 36, add to -36 = 0 (remainder).\n\nSo the quotient is $$a^2 + 4a + 12.$$\n\n8. **Solve the quadratic equation:** $$a^2 + 4a + 12 = 0.$$\nUse the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 48}}{2} = \frac{-4 \pm \sqrt{-32}}{2}.$$\n\n9. **Simplify the discriminant:** $$\sqrt{-32} = \sqrt{-1 \times 32} = i\sqrt{32} = i\sqrt{16 \times 2} = 4i\sqrt{2}.$$\n\n10. **Write the complex roots:** $$a = \frac{-4 \pm 4i\sqrt{2}}{2} = -2 \pm 2i\sqrt{2}.$$\n\n**Final answer:** The solutions are $$a = 3, \quad a = -2 + 2i\sqrt{2}, \quad a = -2 - 2i\sqrt{2}.$$