Question 712235
The unknown being referenced is the width.  


Let w be width
Length is w+1
Height is w-2


{{{w(w+1)(w-2)=40}}}


Some steps,
{{{w(w^2+w-2w-2)-40=0}}}
w(w^2-w-2)-40=0
{{{w^3-w^2-2w-40=0}}}


Solving that is faster if you know synthetic division.  You would use the idea of the Rational Roots Theorem to find values for w to satisfy the cubic equation.  I'd suggest first working with +/-4, +/-5, +/-8, +/-10.   When you get just one first root, the next two will be easy.


Some further work:
Good News!  I tried some long divisions and found +5 or -5 are not roots, but that +4 is a root.  One of the binomials is (w-4).  The quotient from this was {{{w^2+3w+10}}}.  So this means your polynomial equation is
{{{(w-4)(w^2+3w+10)=0}}}, so the quadratic part should be much easier to manage.

NOTE: The discriminant for that quadratic is -31, so the solution will contain an imaginary part.  The only reasonable value for w is 4.