3x^2-2x should be the correct answer
Recall your d = rt, distance = rate * time.
A = first train.
B = second train leaving 2 hrs later.
if say by the time they meet, train B has traveled "t" hours, we know that train A has been traveling 2 hours more than that, because it left 2 hours earlier than train B, thus it has traveled "t + 2" hours.
keeping in mind that by the time they meet, they both have traveled "d" kilometers.
![\bf \begin{array}{lcccl} &\stackrel{km s}{distance}&\stackrel{km/h}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{Train A}&d&75&t\\ \textit{Train B}&d&45&t+2 \end{array} \\\\\\ \begin{cases} d=75t\implies \frac{d}{75}=\boxed{t}\\\\ d=45(t+2)\\ ----------\\ d=45\left( \boxed{\frac{d}{75}}+2 \right) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%0A%26%5Cstackrel%7Bkm%20s%7D%7Bdistance%7D%26%5Cstackrel%7Bkm%2Fh%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%0A%26------%26------%26------%5C%5C%0A%5Ctextit%7BTrain%20A%7D%26d%2675%26t%5C%5C%0A%5Ctextit%7BTrain%20B%7D%26d%2645%26t%2B2%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ad%3D75t%5Cimplies%20%5Cfrac%7Bd%7D%7B75%7D%3D%5Cboxed%7Bt%7D%5C%5C%5C%5C%0Ad%3D45%28t%2B2%29%5C%5C%0A----------%5C%5C%0Ad%3D45%5Cleft%28%20%5Cboxed%7B%5Cfrac%7Bd%7D%7B75%7D%7D%2B2%20%5Cright%29%0A%5Cend%7Bcases%7D)
58/11 = 20/x
cross multiply because this is a proportion
(58)(x) = (11)(20)
58x = 220
x = 220/58
x = 3.79 inches