A rectangular piece of land measures 12m by 5m. A path of constant width runs along one side and one end. If the total area of the plot and the path is 120m², find the width of the path

Let the width of the path be denoted as \( x \). The rectangular piece of land has dimensions of 12 m by 5 m. When the path is added along one side and one end, the dimensions of the new area will change as follows:

- The length is extended by the width of the path on one end:
\[
\text{New length} = 12 + x
\]
- The width is extended by the width of the path on one side:
\[
\text{New width} = 5 + x
\]

The total area of the plot (land + path) can be expressed as the product of the new dimensions:
\[
\text{Total area} = (\text{New length}) \times (\text{New width}) = (12 + x)(5 + x)
\]

According to the problem, the total area is given as 120 m²:
\[
(12 + x)(5 + x) = 120
\]

Now, we will expand the left side of the equation:
\[
12 \cdot 5 + 12x + 5x + x^2 = 120
\]
\[
60 + 17x + x^2 = 120
\]

Next, we will rearrange the equation to bring all terms to one side:
\[
x^2 + 17x + 60 - 120 = 0
\]
\[
x^2 + 17x - 60 = 0
\]

To solve the quadratic equation \( x^2 + 17x - 60 = 0 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = 17, c = -60 \).

First, we calculate the discriminant:
\[
b^2 - 4ac = 17^2 - 4 \cdot 1 \cdot (-60) = 289 + 240 = 529
\]

Now we compute \( x \):
\[
x = \frac{-17 \pm \sqrt{529}}{2 \cdot 1}
\]
\[
\sqrt{529} = 23
\]
\[
x = \frac{-17 \pm 23}{2}
\]

This results in two potential solutions:
\[
x = \frac{6}{2} = 3 \quad \text{(since \( x \) cannot be negative)}
\]
and
\[
x = \frac{-40}{2} = -20 \quad \text{(not feasible)}
\]

Thus, the width of the path is:
\[
\boxed{3 \text{ m}}
\]