<PRE><B>Question 871050</B>
speed to go there	x					
speed returning	x	+	10			
						
Distance = same		300	km   			
original time – time with increased speed =					1	hour
t=d/t						

300	/	x	-	300	/	(	x	+	10	)	=	1				
LCD=	x	(	x	+	10	)										
multiply by LCD																
300	(	x	+	10	)	-	300	x	=	1	x	(	x	+	10	)
300	x	+	3000	-	300	x	=	1	X^2	+	10	x				
3000	=	1	X^2	+	10	x										
1	X^2		+	10	x	-	-3000	=	0							
Find roots of the quadratic equation																
a=	1		b=	10		c=	-3000									
																
x1=	(	-10	+	sqrt(		100	-	12000	))	/	2					
x1=(	-10	+	110	)/	2											
x1=	50															
																
x2=	(	-7	-	sqrt(		49	-	20	)	/	2					
x2=(	-10	-	110	)/	2											
x2=	-60									ignore negative value

to go there speed = 50 	mph

returning speed = 60 mph					
																
</PRE>