1. **State the problem:** Solve the equation $x^2 - x^3 = 12$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation equal to zero: $$x^2 - x^3 - 12 = 0$$ 3. **Rearrange terms:** Write in standard polynomial form: $$-x^3 + x^2 - 12 = 0$$ Or equivalently: $$x^3 - x^2 + 12 = 0$$ (multiplying both sides by $-1$) 4. **Try to find rational roots:** Use the Rational Root Theorem. Possible rational roots are factors of 12 divided by factors of 1, i.e., $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. 5. **Test $x=2$:** $$2^3 - 2^2 + 12 = 8 - 4 + 12 = 16 \neq 0$$ 6. **Test $x=-2$:** $$(-2)^3 - (-2)^2 + 12 = -8 - 4 + 12 = 0$$ So $x = -2$ is a root. 7. **Factor out $(x + 2)$:** Use polynomial division or synthetic division: $$x^3 - x^2 + 12 = (x + 2)(x^2 - 3x + 6)$$ 8. **Solve quadratic $x^2 - 3x + 6 = 0$:** Use quadratic formula: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 24}}{2} = \frac{3 \pm \sqrt{-15}}{2}$$ 9. **Simplify roots:** $$x = \frac{3 \pm i\sqrt{15}}{2}$$ 10. **Final solutions:** $$x = -2, \quad x = \frac{3 + i\sqrt{15}}{2}, \quad x = \frac{3 - i\sqrt{15}}{2}$$ These are the three roots of the equation, one real and two complex conjugates.