We are given the quadratic equation \( w^2 + 6w - 112 = 0 \) to find the width \( w \) of the trampoline.
To solve the quadratic equation, 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 \( b^2 - 4ac \): \[ b^2 = 6^2 = 36 \] \[ 4ac = 4 \times 1 \times (-112) = -448 \] \[ b^2 - 4ac = 36 - (-448) = 36 + 448 = 484 \]
Next, we calculate the square root of the discriminant: \[ \sqrt{484} = 22 \]
Now, we can substitute these values into the quadratic formula: \[ w = \frac{-6 \pm 22}{2 \times 1} = \frac{-6 \pm 22}{2} \]
This gives us two possible solutions for \( w \):
- \( w = \frac{-6 + 22}{2} = \frac{16}{2} = 8 \)
- \( w = \frac{-6 - 22}{2} = \frac{-28}{2} = -14 \)
Since a width cannot be negative, we take the positive solution: \[ w = 8 \]
Thus, the width of the trampoline is \( \boxed{8} \) feet.
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