## Problem\nSolve for $a$ in the equation $a^3+a^2=36$.\n\n1. **Start with the given equation**\n$$a^3+a^2=36$$\n\n2. **Move everything to one side**\n$$a^3+a^2-36=0$$\n\n3. **Factor the cubic**\nTry factoring with integer roots.\nCheck $a=3$: $3^3+3^2=27+9=36$, so $a-3$ is a factor.\nThen\n$$a^3+a^2-36=(a-3)(a^2+4a+12)$$\n\n4. **Set each factor equal to zero**\n$$ (a-3)(a^2+4a+12)=0$$\nSo either\n$$a-3=0$$\nor\n$$a^2+4a+12=0$$\n\n5. **Solve the first factor**\n$$a-3=0\Rightarrow a=3$$\n\n6. **Check the quadratic factor**\n$$a^2+4a+12=0$$\nCompute the discriminant: $\Delta=b^2-4ac=4^2-4(1)(12)=16-48=-32$.\nSince $\Delta<0$, there are **no real** solutions from this quadratic.\n\n7. **Final answer (real solutions)**\n$$a=3$$\n\n8. **Optional note (complex solutions)**\nIf you allow complex numbers, the quadratic gives\n$$a=-2\pm 2i\sqrt{2}$$