Before we do anything, we can factorize the numerator and denominator:
[tex]\dfrac{x^2+9x+20}{x^2-x-20}=\dfrac{(x+5)(x+4)}{(x-5)(x+4)}=\dfrac{x+5}{x-5}[/tex]
(the cancellation is valid as long as [tex]x\neq-4[/tex]; [tex]x=-4[/tex] is not even in the domain of the right hand side, so we're good here)
Then
[tex]\drac{x+5}{x-5}=\dfrac{x-5+10}{x-5}=1-\dfrac{10}{x-5}>0\implies1>\dfrac{10}{x-5}[/tex]
[tex]\implies10<x-5\implies x>15[/tex]
