A rectangle is drawn so the width is 3 inches longer than the height. If the rectangle's diagonal measurement is 46 inches, find the height.

Let the height be = h

Then the width is h + 3

And by the Pythagorean Theorem, we have

h^2 + (h + 3) ^2  = 46^2    simplify

h^2 + h^2 + 6h + 9 = 2116

2h^2 + 6h - 2107  = 0    using the onsite solver we have

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{h}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{h}{\mathtt{\,-\,}}{\mathtt{2\,107}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{h}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{4\,223}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
{\mathtt{h}} = {\frac{\left({\sqrt{{\mathtt{4\,223}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{h}} = -{\mathtt{33.992\: \!306\: \!781\: \!759\: \!894}}\\
{\mathtt{h}} = {\mathtt{30.992\: \!306\: \!781\: \!759\: \!894}}\\
\end{array} \right\}$$

h = about 31 inches   (rounded)