Question 881734
Since it's a right triangle, use the Pythagorean theorem,
{{{x^2+(x+1)^2=(x+2)^2}}}
{{{x^2+x^2+2x+1=x^2+4x+4}}}
{{{x^2-2x-3=0}}}
{{{(x-3)(x+1)=0}}}
Two solution however since it was stated that x>0 only the positive solution works here.
{{{x-3=0}}}
{{{x=3}}}
So the triangle has sides 3,4,5.