so.. hmm notice the picture here
we know the length is 59 more than twice the width,
twice the width, 2 * w, or 2w
59 more than that
2w + 59
we also know that, whatever that is, the diagonal of it is,
2 more inches than that, or
(2w + 59) + 2
now.. use the pythagorean theorem
[tex]\bf c^2=a^2+b^2\implies \qquad
\begin{cases}
a=2w+59\\
b=w\\
c=(2w+59)+2\to 2w+61
\end{cases}
\\\\\\
(2w+61)^2=(2w+59)^2+(w)^2
\\ \qquad\qquad \uparrow\qquad \qquad\qquad \qquad \uparrow \\
\textit{expand using binomial theorem, or FOIL}[/tex]
you'd end up with a quadratic, after simplifying, solve for "w"
