The original equation states that
[tex]300=vt[/tex]
We make the same distance, with higher speed, in less time:
[tex]300=(v+5)(t-2)[/tex]
Comparing the two equations, we deduce
[tex]vt = (v+5)(t-2) = vt-2v+5t-10 \iff -2v+5t-10=0 \iff v = \dfrac{5t-10}{2}[/tex]
If we substitute this expression for v in the first equation, we have
[tex]300=\left(\dfrac{5t-10}{2}\right)t=\dfrac{5t^2-10t}{2}[/tex]
We deduce
[tex]5t^2-10t=600 \iff t^2-2t-120=0[/tex]
whose solutions are
[tex]t=-10,\quad t=12[/tex]
We can only accept the positive solution, so we have t=12. Substitute this again in the first equation:
[tex]300=v\cdot 12 \implies v = \dfrac{300}{12} = 25[/tex]
