Question 871468
using Pythagoras theorem

one leg be x

orther leg = x=2

x^2+(x+2)^2=6^2

x^2+x^2+4x+4=36

2x^2+4x+4=36
2x^2+4x-32=0
/2
x^2+2x-16=0


Find the roots of the equation by quadratic formula								
								
a=	1	,	b=	2	,	c=	-16	
								
b^2-4ac=	4	+	64					
b^2-4ac=	68							
{{{	sqrt(	68	)=	8.25	}}}			
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}								
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}								
x1=(	-2	+	8.25	)/	2			
x1=	3.12							
{{{x2=(-b-sqrt(b^2-4ac))/(2a)}}}								
x2=(	-2	-8.25	) /	2				
x2=	-5.12							

x= 3.12 OR -5.12