Question 32003
Let the length be x
let the width be x+2
THis makes it a pythagorean theorem question:
Equation:
{{{x^2+(x+2)^2=6^2}}}
{{{x^2+x^2+4x+4=36}}}
{{{2x^2+4x-32=0}}}
{{{x^2+2x-16=0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=1, b=2, c=-16
{{{x = (-2 +- sqrt( 2^2-4*1*-16 ))/(2*1) }}}
{{{x=(-2+-sqrt(4+64))/2}}}
Simplfy and remove the negative:
x=3.12
3.12+2=5.12
Hence, the width is 5.12 and the length is 3.12

Paul.