Solve the equation using the quadratic formula. \( x^{2}-4 x+8=0 \) The solution set is (Simplify your answer. Type an exact answer, using radicals and \( i \) as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)

To solve the quadratic equation \( x^{2} - 4x + 8 = 0 \) using the quadratic formula, we start by identifying the coefficients \( a \), \( b \), and \( c \): - \( a = 1 \) - \( b = -4 \) - \( c = 8 \) The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Now, we will calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (-4)^2 - 4(1)(8) = 16 - 32 = -16 \] Since the discriminant is negative, we will have complex solutions. Now we can substitute the values into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{-16}}{2(1)} = \frac{4 \pm \sqrt{-16}}{2} \] Next, we simplify \( \sqrt{-16} \): \[ \sqrt{-16} = 4i \] Now substituting this back into the formula gives: \[ x = \frac{4 \pm 4i}{2} \] We can simplify this further: \[ x = \frac{4}{2} \pm \frac{4i}{2} = 2 \pm 2i \] Thus, the solution set is: \[ \{ 2 + 2i, 2 - 2i \} \] So the final answer is: \[ 2 + 2i, 2 - 2i \]