Question 949117
a=shorter leg; b=longer leg=a+2; c=hypotenuse=a+4
{{{a^2+b^2=c^2}}} Substitute for b and c
{{{a^2+(a+2)^2=(a+4)^2}}}
{{{a^2+a^2+4a+4=a^2+8a+16}}} Subtract a^2 from each side.
{{{a^2+4a+4=8a+16}}} Subtract 8a from each side.
{{{a^2-4a+4=16}}} Subtract 16 from each side.
{{{a^2-4a-12=0}}}
*[invoke quadratic "a", 1, -4, -12 ]
So the solution we want is 6
The length of the shorter leg is 6 mm.
b=a+2 mm=6 mm+2 mm=8 mm  The length of the longer leg is 8 mm.
c=a+4 mm=6 mm+4 mm=10 mm  The length of the hypotenuse is 10 mm, so we have a 3-4-5 right triangle.