Question 858288
a=	1	,	b=	-4	,	c=	4
							
b^2-4ac=	16	+	-16				
b^2-4ac=	0						
{{{	sqrt(	0	)=	0	}}}		
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(	4	+	0	)/	2		
x1=	2						
{{{x2=(-b-sqrt(b^2-4ac))/(2a)}}}							
x2=(	4	0	) /	2			
x2=	2						

The root is 2