Question 486393
A ---- North ---- 5 km/h

B----- east ------16km/h (After 1 hour)

let the time be t 

A distance 5t+5

B distance 16 t

(5t+5)^2+(16t)^2=52^2 ( I have used x in place of t)

25t^2+50t+25+256t^2=52^2
281t^2+50t-2679 =0
Find the roots of the equation by quadratic formula							
							
a=	281	b=	50	c=	-2679		
							
b^2-4ac=	2500	-	3011196				
b^2-4ac=	3013696			{{{sqrt(	3013696	)}}}=	1736
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(	-50	+	1736	)/	562		
x1=	3		3    				
x2=(	-50	-1736	) /	562			
x2= 	-3.18		-3 1/6				
Ignore negative value							
x	=	3 hours

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