Question 433608
let one leg be x
otehr leg = x+2
hypotenuse = 2x

x^2+(x+2)^2=(2x)^2
x^2+x^2+4x+4=4x^2
2x^2-4x-4=0
Find the roots of the equation by quadratic formula			
a=2  b=	-4  c=	-4		
							
b^2-4ac=16	+	32				
b^2-4ac=48			{{{sqrt(48)}}}=	6.93
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(4+6.93)/4		
x1=	2.73						
x2=(4-6.93)/4			
x2=-0.73						
Ignore negative value	

shortleg = 2.73 cm
longer leg = 4.73 cm
hypotenuse = 5.46cm