Option C: 5, −3
$(x-1)^2 = 16 \Rightarrow x-1 = \pm4 \Rightarrow x = 1+4 = 5$ or $x = 1-4 = -3$.
Take the square root of both sides to get $x - 1 = \pm 4$, then solve both cases. Many students mistakenly choose B ($4, -4$) by forgetting the "$-1$" shift. Always solve $(x - a)^2 = k$ as $x = a \pm \sqrt{k}$.