Question 544480
Original speed	x			
Changed speed	x	-	10	
				
Distance = same		300	miles	
original time – time with decreased speed =0.42
t=d/t				

300/(x-10)-300/x= 5/12

300x-300(x-10)= 5*x(x-10)/12

300x-300x+3000=(5x^2-50x)/12

multiply by 12

36000=5x^2-50x
5x^2-50x-36000=0
/5
x*2-10x-7200=0

x^2-90x+80x-7200=0
x(x-90)+80(x-90)=0
(x-90)(x+80=0
x= 90 mph the average speed.