Question

solve x^2-4x-12=0 by completing the square.

Answer 1

(x^2-4x)-12=0
(x^2-4x+16-16)-12=0
(x^2-4x+16)-28=0
(x-4)^2-28=0

Answer 2

\[x^2-4x-12=0\]
\[X^2-4x=12\]
\[(x-2)^2=12+2^2=12+4=16\]
\[(x-2)^2=16\]
\[x-2=4,\text { or } x-2=-4\]
\[x=4,\text { or }x=-2\]

Answer 3

sorry didn't finish

Answer 4

but just solve that after