Solve the equation by completing the square: \( y^{2}-2 y+19=3 \) \( \begin{array}{l}1-15 i, 1+15 i \\ -1-\sqrt{15} i,-1+\sqrt{15} i \\ -15-\sqrt{3} i,-15+\sqrt{3} i \\ 1-\sqrt{15 i} i+\sqrt{15 i} \\ -1+15 i-1-15 i\end{array} \)

To solve the equation \( y^{2} - 2y + 19 = 3 \) by completing the square, we first rearrange the equation: \( y^{2} - 2y + 16 = 0 \). Next, we complete the square for the left side. 1. Take half of the coefficient of \( y \) (which is \(-2\)), square it, and add it inside the equation: \[ y^{2} - 2y + 1 = 1 \implies (y-1)^{2} = 1 \] 2. Now, substitute back into our equation: \[ (y-1)^{2} = 16. \] 3. Now, take the square root of both sides: \[ y-1 = \pm 4. \] 4. Solve for \( y \): \[ y = 1 + 4 = 5 \quad \text{or} \quad y = 1 - 4 = -3. \] Thus, the solutions are \( y = 5 \) and \( y = -3 \). All calculations confirm that neither of the complex-number options provided (like \( 1 - 15i \) or others) are actual solutions to the original equation. These solutions yield \( y \) values corresponding to the real parts since we worked only with real numbers in this equation.