Find all complex solutions of \( 2 x^{2}-x+8=0 \). (If there is more than one solution, separate them with commas.)

To find the complex solutions of the quadratic equation \( 2x^{2}-x+8=0 \), we can use the quadratic formula given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = -1 \), and \( c = 8 \). First, we need to calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (-1)^2 - 4(2)(8) = 1 - 64 = -63 \] Since the discriminant is negative, this indicates that the solutions will be complex. Now substituting into the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{-63}}{2(2)} = \frac{1 \pm \sqrt{-63}}{4} \] We can simplify \(\sqrt{-63}\): \[ \sqrt{-63} = \sqrt{63}i = \sqrt{9 \cdot 7}i = 3\sqrt{7}i \] Now substituting this back into the equation gives us: \[ x = \frac{1 \pm 3\sqrt{7}i}{4} \] Thus, the solutions are: \[ x = \frac{1 + 3\sqrt{7}i}{4}, \quad x = \frac{1 - 3\sqrt{7}i}{4} \] So, the final answers are: \[ \frac{1 + 3\sqrt{7}i}{4}, \frac{1 - 3\sqrt{7}i}{4} \]