Let usual speed of train is `x` km/h.
Let the time taken by the train to cover `360` km distance with normal speed is `t` hour.
Then, `360/x = t->(1)`
When speed is increased by `20` km/h, then,
`360/(x+20) = t-3->(2)`
Subtracting (1) from (2),
`360/(x+20) - 360/x = -3`
`=>360x -360x-7200 = -3x(x+20)`
`=>3x^2+60x-7200 = 0`
`=>x^2+20x-2400 = 0`, which is the required quadratic equation.
Now, we will solve it to find the speed of the train.
`=> x^2+20x-2400 = 0 `
`=>x^2+60x-40x-2400 = 0`
`=>(x+60)(x-40) = 0`
`=> x = -60 and x = 40`
But, `x` can not be negative, so, `x = 40` km/h.
`:.` Usual speed of the train is `40` km/h.
