Question 549665
one leg x
the other leg x+1

hypotenuse = 6m

Pythagoras theorem								
								
(Hyp)^2= (leg1)^2+Leg2^2

6^2=x^2+(x+1)^2
36=x^2+x^2+2x+1
36=2x^2+2x+1
2x^2+2x-35=0

Find the roots of the equation by quadratic formula									
									
a=	2	,b=	2	,c=	-35				
									
b^2-4ac=	4	+	280						
b^2-4ac=	284								
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}									
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}									
{{{x1=(-12+21)/(4)}}}									
x1=(	-2	+	16.85	)/	4				
x1=	3.71								
x2=(	-2	-16.85	) /	4					
x2= 	-4.71