Question 930283

s (mi/hr) the speed  of the slower train
{{{180mi/s  - 180/(s+20) = 1.5hr}}}
180(s+20) - 180s = 1.5s(s+20)
{{{(3/2)s^2 + 30s - 3600 = 0}}} |Multiplying thru by 2/3 so as all denominators = 1
s^2 + 20s - 2400 = 0 (tossing out negative solution for unit measure)
(s+60)(s-40)= 0
s = 40mph, the speed  of the slower train.  Faster train 60mph
And...checking
180mi/40mph - 180mi/60mph = 4.5hr - 3hr = 1.5hr