The information in the problem involves two variables: the usual average speed v, and the usual time t.

so 390 km = v*t and,

390 km = (v + 6)(t - 15/60)

From the first equation, t = 390 / v.

Plugging that into the second equation, we have:

390 = (v + 6)(390/v - 15/60)

390 = 390 + 6*390/v - 15v/60 - 15*6/60

0 = 6*390 - 15v^2/60 - 15*6v/60

Using the quadratic equation, v = (-b +- sqrt(b^2-4ac))/2a

So v = -99.7936 and 93.7936 km/h

Clearly the physical answer in this case is **v = 93.8 km/h,**
since we assumed that v was positive.

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