To find the width of the trampoline, we can solve the quadratic equation given:
\[ w^2 + 6w - 112 = 0. \]
We can use the quadratic formula:
\[ w = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}, \]
where \( a = 1 \), \( b = 6 \), and \( c = -112 \).
First, we calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = 6^2 - 4(1)(-112) = 36 + 448 = 484. \]
Now we can find \( w \):
\[ w = \frac{{-6 \pm \sqrt{484}}}{2(1)}. \]
The square root of 484 is 22:
\[ w = \frac{{-6 \pm 22}}{2}. \]
This gives us two possible solutions:
\[ w = \frac{{-6 + 22}}{2} = \frac{16}{2} = 8, \]
\[ w = \frac{{-6 - 22}}{2} = \frac{-28}{2} = -14. \]
Since the width cannot be negative, we ignore the second solution. Thus, the width of the trampoline is:
\[ w = 8 \text{ ft}. \]
Therefore, the answer is c) 8 ft.