Solve the equation by completing the square: \( y^{2}-10 y+33=3 \) \begin{tabular}{|l}\( 5-\sqrt{5} i, 5+\sqrt{5} i \) \\ \( 5-5 i, 5+5 i \) \\ \( -5-\sqrt{5} i,-5+\sqrt{5} i \) \\ \( -5-\sqrt{3} i,-5+\sqrt{3} i \) \\ \( -5+5 i,-5-5 i \)\end{tabular}

To solve the equation \( y^{2}-10 y + 33 = 3 \), we first simplify it by moving the constant value on the right side: \[ y^{2} - 10y + 30 = 0 \] Next, we complete the square. We take the coefficient of \( y \) (which is -10), divide it by 2 to get -5, and then square it to get 25. We then rewrite the quadratic expression: \[ y^{2} - 10y + 25 = -5 \] This can be factored as: \[ (y - 5)^{2} = -5 \] From here, we take the square root of both sides: \[ y - 5 = \pm \sqrt{-5} \] Recognizing that \( \sqrt{-5} = \sqrt{5}i \), we can find the values for \( y \): \[ y - 5 = \sqrt{5}i \quad \text{or} \quad y - 5 = -\sqrt{5}i \] Solving these gives: \[ y = 5 + \sqrt{5}i \quad \text{and} \quad y = 5 - \sqrt{5}i \] Thus, the solutions to the equation are: \[ y = 5 - \sqrt{5} i, \quad y = 5 + \sqrt{5} i \] Therefore, the correct options from the given list are \( 5 - \sqrt{5} i \) and \( 5 + \sqrt{5} i \).