Quadratic Equations

A journey of $300 \; km$ would take $2$ hours less if the speed was increased by $5 \; kmph$. Find the original speed.

Distance covered $= d = 300 \; km$

Let original speed = $s \; kmph$

Then, original time $= t_1 = \dfrac{d}{s} = \dfrac{300}{s} \;$ hours

New speed $= s + 5 \; kmph$

$\therefore \;$ New time $= t_2 = \dfrac{300}{s + 5} \;$ hours

As per sum, $\;$ $t_2 = t_1 - 2$

i.e. $\;$ $\dfrac{300}{s + 5} = \dfrac{300}{s} - 2$

i.e. $\;$ $\dfrac{150}{s + 5} = \dfrac{150}{s} - 1$

i.e. $\;$ $150 s = 150 s + 750 - s \left(s + 5\right)$

i.e. $\;$ $s^2 + 5 s - 750 = 0$

i.e. $\;$ $s^2 + 30 s - 25 s - 750 = 0$

i.e. $\;$ $s \left(s + 30\right) - 25 \left(s + 30\right) = 0$

i.e. $\;$ $\left(s + 30\right) \left(s - 25\right)= 0$

i.e. $\;$ $s = -30$ $\;$ or $\;$ $s = 25$

$\because \;$ Original speed cannot be negative, original speed $= 25 \; kmph$