The original speed is 1/t mi/h. The increased speed is 1/t + 5 mi/h. The time to go one mile at the increased speed is given by: $\large \frac{1}{\frac{1}{t}+5}\ .......(1)$ We can equate the time in (1) with the decreased time to go one mile as follows: $\large \frac{1}{\frac{1}{t}+5}=t-\frac{12}{3600}\ .........(2)$ Manipulation of (2) produces the following quadratic equation: $\large 1500t ^{2}-5t-1=0\ ...........(3)$ The real solution of (3) gives the original time to travel one mile as a decimal fraction of an hour. The reciprocal gives the original speed in mi/h.