1. **State the problem:** Solve the cubic equation $$2x^3 - 3x^2 - 9x + 10 = 0$$ for $x$. 2. **Recall the approach:** For cubic equations, we can try to find rational roots using the Rational Root Theorem, then factor the polynomial. 3. **Possible rational roots:** Factors of constant term 10 are $\pm1, \pm2, \pm5, \pm10$. Factors of leading coefficient 2 are $\pm1, \pm2$. Possible roots are $\pm1, \pm\frac{1}{2}, \pm2, \pm5, \pm\frac{5}{2}, \pm10$. 4. **Test $x=1$:** $$2(1)^3 - 3(1)^2 - 9(1) + 10 = 2 - 3 - 9 + 10 = 0$$ So, $x=1$ is a root. 5. **Divide polynomial by $(x-1)$:** Using polynomial division or synthetic division: $$\frac{2x^3 - 3x^2 - 9x + 10}{x - 1} = 2x^2 - x - 10$$ 6. **Solve quadratic $2x^2 - x - 10 = 0$:** Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-10)}}{2(2)} = \frac{1 \pm \sqrt{1 + 80}}{4} = \frac{1 \pm \sqrt{81}}{4}$$ 7. **Simplify roots:** $$x = \frac{1 \pm 9}{4}$$ So, $$x = \frac{1 + 9}{4} = \frac{10}{4} = \frac{5}{2}$$ $$x = \frac{1 - 9}{4} = \frac{-8}{4} = -2$$ 8. **Final solutions:** $$x = 1, \quad x = \frac{5}{2}, \quad x = -2$$