Find the zeros, if any, of the quadratic function using the quadratic formula. What are the \( x \)-intercepts, if any, of the graph of the function? \( f(x)=2 x^{2}+3+6 x \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. The zeros and the \( x \)-intercepts are different. The zeros are \( \square \), the \( x \)-intercepts are B. The zeros and the x-intercepts are the same. They are C. There is no real zero solution and no \( x \)-intercept.

To find the zeros of the quadratic function \( f(x) = 2x^2 + 6x + 3 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, the coefficients are: - \( a = 2 \) - \( b = 6 \) - \( c = 3 \) Now, we can calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = 6^2 - 4 \cdot 2 \cdot 3 = 36 - 24 = 12 \] Since the discriminant is positive, there will be two distinct real zeros. Now we can substitute the values into the quadratic formula: \[ x = \frac{-6 \pm \sqrt{12}}{2 \cdot 2} \] We can simplify \( \sqrt{12} \): \[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \] Now substituting this back into the formula: \[ x = \frac{-6 \pm 2\sqrt{3}}{4} \] This simplifies to: \[ x = \frac{-3 \pm \sqrt{3}}{2} \] Thus, the two zeros are: \[ x_1 = \frac{-3 + \sqrt{3}}{2}, \quad x_2 = \frac{-3 - \sqrt{3}}{2} \] Now, we can summarize the results: A. The zeros and the \( x \)-intercepts are different. The zeros are \( \frac{-3 + \sqrt{3}}{2}, \frac{-3 - \sqrt{3}}{2} \), the \( x \)-intercepts are the same as the zeros. So the correct choice is: **B. The zeros and the x-intercepts are the same. They are \( \frac{-3 + \sqrt{3}}{2}, \frac{-3 - \sqrt{3}}{2} \).**