Question 942809
So the legs would be {{{x}}} and {{{x-2}}} and the hypotenuse would be {{{x+2}}}.
Use the Pythagorean theorem,
{{{(x+2)^2=x^2+(x-2)^2}}}
{{{x^2+4x+4=x^2+x^2-4x+4}}}
{{{x^2+4x+4=2x^2-4x+4}}}
{{{x^2-8x=0}}}
{{{x(x-8)=0}}}
Only non-zero solutions make sense here,
{{{x-8=0}}}
{{{x=8}}}
So the lengths are 6,8, and 10.