Question 579635
You can check that 7 and 12 works and you're done.


Or, you can do


*[tex \LARGE h(h+5) = 84 \Rightarrow h^2 + 5h - 84 = 0]


The LHS factors:


*[tex \LARGE (h-7)(h+12) = 0]


h = 7 or h = -12. We discard the negative solution since length can't be negative, so h = 7.