Answer
$x \in (-\infty, 1) \cup (2, \infty )$
Work Step by Step
$H(x)=\log_{2} (x^2-3x+2)$
Domain of $\log x$ is $x >0$.
Thus, to find the domain of $\log_{2} (x^2-3x+2),$
$x^2-3x+2>0,$
$x^2-x-2x+2>0,$
$x(x-1)-2(x-1)>0$
$(x-2)(x-1)>0,$
$x<1$ or $x>2$.
Therefore, The domain is $x \in (-\infty, 1) \cup (2, \infty )$
