If h=-16t^2+48t-20 and if h>12 you plug >12 into the h position in the equation:
12<-16t^2+48t-20; simplified to: 0<-16(t^2-3t+2)
Then you solve THIS equation by factoring and you should get:
0<(t-1)(t-2). 1<t and 2<t are your t(x) values. So when h(y)>12, t>1 & 2.
I hope this helps. You didn't give any answer choices so I'm not sure how they want the answer to look.