Question 246443
right triangle
c^2=a^2+b^2
where c is the hypotenuse and a and b are the legs.

c=a+1
solve for a
a=c-1
c=b+8
solve for b
b=c-8
substitute
c^2=(c-1)^2+(c-8)^2
(c-1)^2=c^2-2c+1
(c-8)^2=c^2-16+64
c^2=2c^2-18c+65
subtract c^2
c^2-18c+65=0
13 and 5
13,12, 5
5,4,-3
but -3 can't be a side of a triangle
so we have 13, 12, and 5
13 is hypotenuse
12 and 5 are the legs



*[invoke quadratic "x", 1, -18, 65 ]