Question 453406
let hypotenuse be x

the legs are (x-5)
By pythagoras theorem

(x-5)^2+(x-5)^2=x^2
x^2-10x+25+x^2-10x+25=x^2
x^2-20x+50=0

Find the roots of the equation by quadratic formula							
							
a=	1	,b=	-20	,c=	50		
							
b^2-4ac=400-200				
b^2-4ac=200
{{{sqrt(200)}}}=14.14
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(20+14.14)/2		
x1=	17.07						
x2=(20-14.14)/2			
x2= 	2.93						
x2 is not possible
hyptenuse = 17.07