Let the width of the garden be \( w \) feet. Then the length will be \( w + 3 \) feet.
According to the problem, the area of the garden is given by the equation:
\[ \text{Area} = \text{length} \times \text{width} = (w + 3) \times w = 180 \]
Expanding this gives us:
\[ w^2 + 3w = 180 \]
Rearranging the equation, we have:
\[ w^2 + 3w - 180 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), and \( c = -180 \).
Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-180) = 9 + 720 = 729 \]
Now applying this to the quadratic formula:
\[ w = \frac{-3 \pm \sqrt{729}}{2 \cdot 1} = \frac{-3 \pm 27}{2} \]
This results in two potential solutions:
- \( w = \frac{24}{2} = 12 \) (valid since width can't be negative)
- \( w = \frac{-30}{2} = -15 \) (not valid)
So, the width \( w = 12 \) feet. The length will then be:
\[ l = w + 3 = 12 + 3 = 15 \text{ feet} \]
Thus, the dimensions of the vegetable garden should be 12 feet by 15 feet.
Among the responses you provided, the correct answer is 12 feet by 15 feet.