### 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. First try to split the polynomial using a common factor pattern\nWe look for an integer $r$ such that $(a-r)$ is a factor. Test $a=3$: $3^3+3^2-36=27+9-36=0$, so $(a-3)$ is a factor.\n\n4. Divide (or factor) to get the full factorization\n$$a^3+a^2-36=(a-3)(a^2+4a+12)$$\n\n5. Check the quadratic for real roots\nSet the quadratic factor to zero\n$$a^2+4a+12=0$$\nDiscriminant\n$$\Delta=4^2-4\cdot1\cdot12=16-48=-32$$\nSo there are no real solutions from this factor.\n\n6. Solve the linear factor\n$$a-3=0$$\n$$a=3$$\n\n### Final Answer\n$a=3$ (real solution).