Let the original speed of the train = x km/h.
Distance = 90 km.
Time taken at original speed = $\dfrac{90}{x}$ hours.
Time taken at increased speed = $\dfrac{90}{x+15}$ hours.
Since the increased speed reduces time by 30 minutes $= \dfrac{1}{2}$ hour:
$$\frac{90}{x} - \frac{90}{x+15} = \frac{1}{2}$$
$$90(x+15) - 90x = \frac{x(x+15)}{2}$$
$$90 \times 15 = \frac{x(x+15)}{2}$$
$$1350 \times 2 = x^2 + 15x$$
$$x^2 + 15x - 2700 = 0$$
Factorising:
$$x^2 + 60x - 45x - 2700 = 0$$
$$x(x + 60) - 45(x + 60) = 0$$
$$(x - 45)(x + 60) = 0$$
So, $x = 45$ or $x = -60$.
Since speed cannot be negative, $x = -60$ is rejected.
∴ The original speed of the train = 45 km/h.
Source: Chapter 4 (Quadratic Equations), Exercise 4.2
---