Question 150354
In a  rectangle, the one side is always longer than the other.
We mark "x"= height.
And the other side is 2m longer than this, so "x+2m"=the longer side (accdg. to your problem is the width).
Since we have a brace of {{{sqrt(6)}}}, go by Pyth. Theorem to solve the other sides.
{{{(sqrt(6))^2=x^2+(x+2)^2}}}
{{{6=x^2+x^2+4x+4}}} -----------> {{{2x^2+4x-2=0}}}----> {{{x^2+2x-1=0}}}
By quadratic,
*[invoke quadratic "x", 1, 2, -1]
Use {{{x=0.414m}}}
To check,
{{{(sqrt(6))^2=0.414^2+(2+0.414)^2}}}
{{{6=0.17+5.83}}}
{{{6m=6m}}}
Thank you,
Jojo