Question 1116617
If x is longer leg, x-2 is the shorter leg, then

{{{x^2+(x-2)^2=10^2}}}

{{{x^2+x^2-4x+4-100=0}}}

{{{2x^2-4x-96=0}}}

{{{x^2-2x-48=0}}}

{{{(x+6)(x-8)=0}}}


{{{highlight(x=8)}}}-------longer leg


{{{highlight(6)}}}---------shorter leg