Answer:
[tex]\large\boxed{h>2\ \vee\ h<1\to h\in(-\infty,\ 1)\ \cup\ (2,\ \infty)}[/tex]
Step-by-step explanation:
[tex]|2h-3|>1\iff2h-3>1\ \vee\ 2h-3<-1\qquad\text{add 3 to both sides}\\\\2h>4\ \vee\ 2h<2\qquad\text{divide both sides by 2}\\\\h>2\ \vee\ h<1\to h\in(-\infty,\ 1)\ \cup\ (2,\ \infty)[/tex]
