140= (v+5) (t-0.5) The formula is vt=140, solve for the vaiable "v"

Question

A bakery driver typically leaves the bakery at 5 a.m. each morning on his 140 mile route.  One day he gets a late start and unable to leave until 5:30.  To finish the route on time he drives 5 miles per hour faster than usual.  What speed does he usually drive?

Answer 1

If his normal speed is v mph then the distance in miles is x=vt where t is the time in hours to complete the journey. Since x=140, vt=140, so t=140/v hours.

If his speed changes to v=v+5, then (v+5)(140/v-0.5)=140, because the time is reduced by 0.5 hours.

140-0.5v+700/v-2.5=140,

-0.5v+700/v-2.5=0,

-0.5v²+700-2.5v=0.

Multiply through by -2:

v²+5v-1400=0=(v-35)(v+40)

v=35mph.

CHECK 

Journey time = 140/35=4hr.

5mph more than usual=40mph. Time to travel 140 miles=140/40=3.5hr, which is half an hour shorter than normal journey time.