Let w and l be the dimensions (width and length, respectively) of the coop.
We know that the length of the coop is 4 feet less than twice the width, which means that
![l=2w-4](https://img.qammunity.org/2020/formulas/mathematics/college/mnld9n9nvbdu3njlq889cbfaaa8zi5yky2.png)
Also, the area is 510, but the area is the product of the dimensions, so we have
![lw=510](https://img.qammunity.org/2020/formulas/mathematics/high-school/kw0j1qy6l8md27e1oi0znbhd3fvigxnguc.png)
Plug the expression for l in the formula for the area:
![lw=(2w-4)w=510 \iff 2w^2-4w-510=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/7lgrzxmo57eel5mxgjcuplni0v448180x1.png)
We can divide the whole expression by 2 and solve it with the quadratic formula:
![w^2-2w-255=0 \iff w=(2\pm√(4+1020))/(2)=(2\pm 32)/(2)=1\pm 16](https://img.qammunity.org/2020/formulas/mathematics/high-school/vh2bi1bi3in6wr62r4muvta6p39alk585r.png)
So, the two solutions are
![w_1=1-16=-15,\quad w_2=1+16=17](https://img.qammunity.org/2020/formulas/mathematics/high-school/6k5b6smfbk4pb06ah3w1tu4nb7nbktdk0u.png)
The negative solution makes no sense (we can't have negative lengths), so the width must be 17.
We conclude that the length is
![l=2w-4=2\cdot 17-4=34-4=30](https://img.qammunity.org/2020/formulas/mathematics/high-school/gf22jfj4epz27j834jb1x7asprzf8ov4mu.png)