Question

Solve (y-4)^2+63=0 where y is a real number.

Answer 1

Answer:

$y = 4 + 3 \sqrt{7} i$

Step-by-step explanation:

$(y-4)^2+63=0 \\ \\ (y-4)^2 = - 63 \\ \\ (y-4)= \sqrt{ - 63} \\ \\ y - 4 = 3 \sqrt{7} i \\ \\ y = 4 + 3 \sqrt{7} i$