Let s represent the train's original speed.
time = distance / speed
150/s = 150/(s +5) +1 . . . . . . . the original journey took 1 hour longer than at the faster speed
Multiply by s(s+5).
150(s +5) = 150s +s(s+5)
s^2 +5s -750 = 0 . . . . . . . rearrange to standard form
(s +30)(s -25) = 0 . . . . . . . factor
Solutions to this are values of s that make the factors zero.
s = -30 or +25 . . . . . . . . . only the positive solution makes sense
The train's original speed is 25 mph.
