How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$ I have tried doing factoring and completing the square and also AM-GM
$(2-y)(x-1)=-8$
$xy=2x+y+6≥\sqrt{12xy} $

AM-GM works perfectly fine with a small trick.
$2x+y+6=xy \Rightarrow y(x-1)=2x+6 \Rightarrow y = \dfrac{2x+6}{x-1} \Rightarrow xy =\dfrac{2x^2+6x}{x-1}$.
As @Yves Daoust points out $\dfrac{2x^2+6x}{x-1} = 2x+8+ \dfrac{8}{x-1} = 2(x-1)+10+\dfrac{8}{x-1} $.
But by AM-GM,  $ 2(x-1)+10+\dfrac{8}{x-1} \geq 10 + 2\left(\sqrt{2(x-1) \cdot \dfrac{8}{x-1} } \right)  = 10+2\sqrt{16}=18$. This is indeed the minimum of $xy$.
