Question 447410
x+y=3
x^2+y^2=369

x=3-y
substitute the x in eq 2
(3-y)^2+y^2=369
9-6y+y^2+y^2=369
2y^2-6y-360=0
/2
y^2-3y-180=0
Find the roots of the equation by quadratic formula			
a=1,	b=-3,	c=-180	
						
b^2-4ac=	9-(-720)			
b^2-4ac=	729	
{{{sqrt(	729)}}}=27
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}						
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}						
x1=(3+27	)/2	
x1=15					
x2=(3-27	)/2		
x2= -12					
(15 & -12)
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