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, which is:
\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \(a = 1\), \(b = 6\), and \(c = -112\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-112) = 36 + 448 = 484. \]
Next, we take the square root of the discriminant:
\[ \sqrt{484} = 22. \]
Now we can substitute back into the quadratic formula:
\[ w = \frac{-6 \pm 22}{2 \cdot 1} = \frac{-6 \pm 22}{2}. \]
This results in two possible solutions:
\[ w = \frac{-6 + 22}{2} = \frac{16}{2} = 8, \]
\[ w = \frac{-6 - 22}{2} = \frac{-28}{2} = -14 \quad (\text{not valid, since width can't be negative}). \]
So the width of the trampoline is:
\[ \boxed{8 \text{ ft}}. \]
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