Question 69074
Let x= speed of old ones

x+100= speed of new ones

distance (d)=rate(r) times time (t) or d=rt and t=d/r

2800/x= time it took old planes to make the trip

2800/(x+100)= time it takes the new planes to make the trip

Now we are told that the new planes make the trip 30 min (0.5 hr) faster than the old planes.  So our equation to solve is:


2800/x=(2800/(x+100))+0.5   multiply both sides by x(x+100)

2800(x+100)=2800x+0.5x(x+100)  clear parens

2800x+280000=2800x+0.5x^2+50x  subtract 2800x and also 280000 from both sides

2800x-2800x+280000-280000=2800x-2800x-280000+0.5x^2+50x  collect like terms

0=0.5x^2+50x-280000 or  multiply both sides by 2

x^2+100x-560000=0   Quadratic in standard form

This quadratic can be factored:

(x+800)(x-700)=0

discount negative value for speed

So 
x=700 km/hr ------------------------------speed of old planes

x+100=800 km/hr speed of new planes

Ck

2800/x=4 hrs 
2800/(x+100)=3.5 hrs-----30 min less ---checks



Hope this helps----ptaylor