Answer
$20$ feet
Work Step by Step
We use the $x$-coordinates of the points where the entrance meets the ground to find the width of the entrance at ground level. At ground level, $y=0$. So we substitute $0$ for $y$ and solve for $x$:
$$\begin{align*}
y&=-\frac{11}{50}(x-4)(x-24)&&\text{Write equation.}\\
0&=-\frac{11}{50}(x-4)(x-24)&&\text{Substitute }0\text{ for }y\\
0&=(x-4)(x-24)&&\text{Multiply each side by }-\frac{50}{11}\\
x-4&=0\text{ or }x-24=0&&\text{Zero-Product Property.}\\
x&=4\text{ or }x=24&&\text{Solve for }x.
\end{align*}$$ The width of the entrance is the distance between the $x$-coordinates, $4$ and $24$. So the width of the entrance at ground level is $|24-4|=20$ feet.
