\[ \begin{array}{l}\text { Find the real solutions, if any, using the quadratic formula. } \\ 3 x^{2}-13 x+14=0\end{array} \] Select the correct choice below and fill in any answer boxes in your choice. A. The solution set is \( \left\{\begin{array}{l}\text { B. The solution is not a real number. } \\ \text { (Simplify your answer. Use a comma to separate answers as needed.) }\end{array}\right. \)

To find the real solutions to the quadratic equation \( 3x^2 - 13x + 14 = 0 \) using the quadratic formula, follow these steps: **Quadratic Formula:** \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] **Given Equation:** \[ 3x^2 - 13x + 14 = 0 \] Here, \( a = 3 \), \( b = -13 \), and \( c = 14 \). **Step 1: Calculate the Discriminant (\( \Delta \)):** \[ \Delta = b^2 - 4ac \] \[ \Delta = (-13)^2 - 4(3)(14) \] \[ \Delta = 169 - 168 \] \[ \Delta = 1 \] Since the discriminant is positive (\( \Delta > 0 \)), there are two distinct real solutions. **Step 2: Apply the Quadratic Formula:** \[ x = \frac{-(-13) \pm \sqrt{1}}{2 \times 3} \] \[ x = \frac{13 \pm 1}{6} \] **Solution 1:** \[ x = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3} \] **Solution 2:** \[ x = \frac{13 - 1}{6} = \frac{12}{6} = 2 \] **Final Answer:** **A. The solution set is \( 2, \frac{7}{3} \).**