Suppose the area of a rectangle is $73.6 \mathrm{ft}^{2}$ and the length is $8.3 \mathrm{ft}$ longer than the width. Find the length and width in feet of the rectangle.

Enter the length followed by a comma followed by the width. Do NOT enter any equations. If any number in the

Step 1 ：Given that the area of a rectangle is $73.6 \mathrm{ft}^{2}$ and the length is $8.3 \mathrm{ft}$ longer than the width, we need to find the length and width of the rectangle.

Step 2 ：We can set up the equation for the area of a rectangle, which is length times width. In this case, the length is $8.3 \mathrm{ft}$ longer than the width, so we can express the length as $w + 8.3$ where $w$ is the width.

Step 3 ：Substituting this into the area equation gives us $w(w + 8.3) = 73.6$.

Step 4 ：This is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 1$, $b = -8.3$, and $c = -73.6$.

Step 5 ：We can solve this equation using the quadratic formula, which is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 6 ：Substituting the values of $a$, $b$, and $c$ into the quadratic formula gives us two solutions, $x1 = 13.680083945065752$ and $x2 = -5.380083945065751$.

Step 7 ：Since the width cannot be negative, we discard $x2$ and take $x1$ as the width of the rectangle.

Step 8 ：The width of the rectangle is approximately $13.68 \mathrm{ft}$.

Step 9 ：Substituting the width into the equation for the length gives us $l = w + 8.3 = 13.68 + 8.3 = 21.98008394506575$.

Step 10 ：The length of the rectangle is approximately $21.98 \mathrm{ft}$.

Step 11 ：Final Answer: The width of the rectangle is approximately $13.68 \mathrm{ft}$ and the length is approximately $21.98 \mathrm{ft}$. So, the answer is \(\boxed{(13.68, 21.98)}\).