we can write 2 equations


where x is the wide and y the long
we can replace y=x+12 from the second equation on the first

and solve x

factor ussing
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
where a is 1, b is 12 and c -92
replacing
![\begin{gathered} x=\frac{-(12)\pm\sqrt[]{12^2-4(1)(-92)}}{2(1)} \\ \\ x=\frac{-12\pm16\sqrt[]{2}}{2} \\ \\ x=-6\pm8\sqrt[]{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r4l6ecwqii8u173sblfvpgtnvko2i2171p.png)
the two solutions are

the solution must be positive because it is a measure
so x=5.31feet
now we can replace the value of x on any equation to solve y(I will replace on the second equation)

so the measurements are x=5.31 and y=17.31