Question

completing the square
x^2 - 8x + 32 = 0

Answer 1

(x-4)^2-32=0
(x-4)^2=32
x=4+-4*sqrt(2)

Answer 2

\[x^2-8x=-32\]
\[(x-4)^2=-32+16\]
\[(x-4)=-16\]
\[x-4=\pm\sqrt{-16}\]
\[x-4=\pm4i\]
\[x=4\pm4i\]

Answer 3

typo in line 3 it should read
\[(x-4)^2=-16\] sorry