quadratic reduction problem A train is travelling between two stations that are $100$ km apart at a speed
of $v$ km/h.


*

*Express the time taken for the journey in terms of $v$.
Here I got $\ t=\dfrac{100}{v}$.


On the return journey, due to work on the line, the train’s speed is reduced
by $5$ km/h.


*Express the time taken for the return journey in terms of $v$.
Here I got $\ t = \dfrac{100}{v-5}$.


The time taken for the round trip is $4$ hours.


*Express the time for the round trip as an equation in terms of $v$.
Here I went $\ t= \dfrac{100}{v} + \dfrac{100}{v-5}\ $ to give me $\ 100\left(\dfrac{1}{v} + \dfrac{1}{v+5}\right)$.

*Use algebra to show that the equation in $\,(3)\,$ reduces to the quadratic
equation
$$v^2 -55v + 125 = 0.$$
This is where im stuck
 A: First of all on step three, you put
$100\left(\frac 1v + \frac 1{v+5}\right)$ when it should've been $100\left(\frac 1v + \frac 1{v-5}\right)$.
Anyways, going from $100\left(\frac 1v + \frac 1{v-5}\right) = 4$, we get
$\frac {100}v + \frac {100}{v-5} = 4$
We then subtract both sides by $4$ to get $ \frac {100}v + \frac {100}{v-5} - 4 = 0$
Then we multiply both sides by $v(v-5)$ and we get
$100v - 500 + 100v - 4v^2 + 20v = 0$
Next, we add the like terms to get $-4v^2 + 220v - 500 = 0$
Finally, we simplify by dividing both sides by $-4$ and we get
$v^2 - 55v + 125 = 0$
A: Note that in the last formula in item (3) you have probably made a typo. 
I believe it should be $\ 100\left(\dfrac{1}{v} + \dfrac{1}{v-5}\right)$ rather than $\ 100\left(\dfrac{1}{v} + \dfrac{1}{v+5}\right)$.    
Additionally, you may want to writ out the fool statement of (3) explicitly. 
Do not loose any information you are given in the text. More specifically, you did not reflect in your equation the fact that the total duration of round trip is $4$ hours:
$$
t_\text{total} = 100\left(\dfrac{1}{v} + \dfrac{1}{v-5}\right) = 4. \label{*}\tag{*}
$$
Having equation $\eqref{*}$ written out explicitly, it is easy to write the answer for item (4):
$$
\begin{aligned}
100\left(\dfrac{1}{v} + \dfrac{1}{v-5}\right) = 4 
&\implies 
\dfrac{1}{v} + \dfrac{1}{v-5} = \frac{1}{25}   
\\&\implies 
\dfrac{(v-5) + v}{v(v-5)} =  \frac{1}{25}  
\\&\implies 
25\big((v-5) + v \big) = v(v-5)  
\\&\implies 
50 v - 125 = v^2 - 5 v  
\\&\implies 
\bbox[5px, border:3px solid #FF0000]{v^2 - 55v + 125 = 0\,}
\end{aligned}
$$
Q.E.D.
