1. Problem statement: A train travels 240 km but in bad weather its speed is reduced by 20 km/h so it took 2 hours longer to reach the destination than usual; find the normal speed. 2. Let the normal speed be $v$ km/h. 3. We use the relation Distance = Speed \times Time, so Time = Distance/Speed. 4. The usual time is $\frac{240}{v}$ hours, and the time in bad weather is $\frac{240}{v-20}$ hours. 5. According to the problem, $\frac{240}{v-20} = \frac{240}{v} + 2$. 6. Multiply both sides by $v(v-20)$ to clear denominators: $240v - 240(v-20) = 2v(v-20)$. 7. Simplify the equation to get $4800 = 2v^2 - 40v$. 8. Divide by 2 and bring all terms to one side: $v^2 - 20v - 2400 = 0$. 9. Solve with the quadratic formula: $v = \frac{20 \pm \sqrt{20^2 - 4(1)(-2400)}}{2}$. 10. Compute the discriminant and simplify: $v = \frac{20 \pm \sqrt{10000}}{2} = \frac{20 \pm 100}{2}$. 11. The two roots are $v = 60$ and $v = -40$, and we discard the negative root because speed cannot be negative. 12. Final answer: The normal speed is 60 km/h.