1. **State the problem:** Find the solution set of the cubic equation $$2x^3 - 7x^2 + 2x + 3 = 0$$. 2. **Recall the problem:** We want to find the roots of the cubic polynomial, i.e., values of $x$ that satisfy the equation. 3. **Try possible rational roots using Rational Root Theorem:** Possible roots are factors of constant term 3 divided by factors of leading coefficient 2, i.e., $\pm1, \pm\frac{1}{2}, \pm3, \pm\frac{3}{2}$. 4. **Test $x = -1$:** $$2(-1)^3 - 7(-1)^2 + 2(-1) + 3 = 2(-1) - 7(1) - 2 + 3 = -2 - 7 - 2 + 3 = -8 \neq 0$$ 5. **Test $x = 1$:** $$2(1)^3 - 7(1)^2 + 2(1) + 3 = 2 - 7 + 2 + 3 = 0$$ So, $x=1$ is a root. 6. **Divide polynomial by $(x-1)$ using synthetic division:** Coefficients: 2, -7, 2, 3 Bring down 2. Multiply 2*1=2, add to -7: -7+2 = -5. Multiply -5*1 = -5, add to 2: 2 + (-5) = -3. Multiply -3*1 = -3, add to 3: 3 + (-3) = 0. So quotient polynomial is $$2x^2 - 5x - 3$$. 7. **Solve quadratic $2x^2 - 5x - 3 = 0$ using quadratic formula:** $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$$ 8. **Calculate roots:** - $$x = \frac{5 + 7}{4} = \frac{12}{4} = 3$$ - $$x = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$$ 9. **Check which roots are in the options:** Roots are $\{1, 3, -\frac{1}{2}\}$. None of the options include $-\frac{1}{2}$, but option B is $\{-1, 3, 2\}$, option D is $\{-3, 1, 2\}$, option A is $\{-1, 2, 13\}$, option C is $\{-3, -2, 1\}$. Since $-\frac{1}{2}$ is not in any option, check if $x=2$ or $x=-1$ is root: Test $x=2$: $$2(2)^3 - 7(2)^2 + 2(2) + 3 = 2(8) - 7(4) + 4 + 3 = 16 - 28 + 7 = -5 \neq 0$$ Test $x=-1$ (already tested, not root). Test $x=-3$: $$2(-3)^3 - 7(-3)^2 + 2(-3) + 3 = 2(-27) - 7(9) - 6 + 3 = -54 - 63 - 3 = -120 \neq 0$$ Test $x=13$: Very large, unlikely root. Test $x=-2$: $$2(-2)^3 - 7(-2)^2 + 2(-2) + 3 = 2(-8) - 7(4) - 4 + 3 = -16 - 28 - 1 = -45 \neq 0$$ So only roots are $1, 3, -\frac{1}{2}$. Since $-\frac{1}{2}$ is not in options, the closest set with roots $1, 3, 2$ is option B, but $2$ is not root. Therefore, the correct roots are $\boxed{\left\{1, 3, -\frac{1}{2}\right\}}$ which is not listed. **Conclusion:** None of the given options exactly match the solution set. **Final answer:** Roots are $\boxed{\left\{1, 3, -\frac{1}{2}\right\}}$.