We can remove the absolute value sign by splitting the expression into more than one range according to when and when it doesn't change the sign of what's inside.
|t/2+4|+3>2t
First case - positive
If t/2+4>=0, then the inequality asks for t/2+4+3>2t, which rearranges to -3/2t> -7. Divide through by the t coefficient, noticing that the change of sign needs a matching switch of inequality direction:
t<14/3.
When t/2+4>=0, t>=-8. So for this first case we have specified a range of possible t values:
-8<=t<14/3.
Second case - negative
If t/2+4<0, then the inequality asks for -t/2-4+3>2t, which rearranges to -5/2t> 1. Divide through by the t coefficient, noticing that the change of sign needs a matching switch of inequality direction:
t<-2/5.
When t/2+4<0, t<-8. So for this second case we've specified two conditions that both point in the same direction. For them both to hold we take the strongest one, t<-8.
Combine cases
We now have two ranges of t that fit the inequality, t<-8 and -8<=t<14/3. In general it is perfectly acceptable to finish an answer to such a question by finding multiple ranges of validity. However, in this case, the two ranges can be combined because they touch each other at one end. So our final answer is:
t<14/3
Sanity check the answer by comparing graphs of both sides of the inequality:
graph{(y-(|x/2+4|+3))(y-2x)=0 [-40, 40, -20, 20]}