Question 930748
Well let's set this up as a system of equations.

Allow "w" to be the width.
Allow "h" to be the height.
Allow "d" to be the diagonal.

We know from the Pythagorean Theorem that...
{{{a^2 + b^2 = c^2}}}
but in this case
{{{a=w}}}
{{{b=h}}}
{{{c=d=58}}}, so we can say that...

{{{w^2 + h^2 = d^2 = 58^2}}}. Which is our first equation.
Our second equation comes from the "width is 2 inches longer than the height", which means that {{{w = h + 2}}}.

Substituting that into the first equation we can solve using algebra to find the height.

{{{w^2 + h^2 = 58^2}}}
{{{(h + 2)^2 + h^2 = 58^2}}}
{{{h^2 + 4h + 4 + h^2 = 58^2}}}
{{{2h^2 + 4h + 4 = 58^2}}}
{{{2h^2 + 4h - 3360 = 0}}}
{{{h^2 + 2h - 1680 = 0}}}
*[invoke quadratic "h", 1, 2, -1680 ]

Rounding this to one decimal place, we get our final answer to be 40 inches.
Note that we ignore the negative result from the quadratic formula because you can't have a negative distance.