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Problem:

A rectangular garden has a length of 20 meters and a width of 15 meters. A path of uniform width is to be constructed around the garden, increasing the total area to 600 square meters. What is the width of the path?

---

Solution:

#### Step 1: Define the variables
Let the width of the path be \( x \) meters.

#### Step 2: Determine the dimensions of the entire area (garden + path)
- The original length of the garden is 20 meters.
- The original width of the garden is 15 meters.
- Since the path surrounds the garden uniformly, the total length of the entire area (including the path) will be:
\[
20 + 2x
\]
(We add \( x \) on both sides of the length.)
- Similarly, the total width of the entire area will be:
\[
15 + 2x
\]
(We add \( x \) on both sides of the width.)

#### Step 3: Express the total area
The total area of the garden plus the path is given as 600 square meters. The area of a rectangle is calculated as:
\[
\text{Area} = \text{Length} \times \text{Width}
\]
Thus, the total area can be expressed as:
\[
(20 + 2x)(15 + 2x) = 600
\]

#### Step 4: Expand the equation
Expand the left-hand side of the equation:
\[
(20 + 2x)(15 + 2x) = 20 \cdot 15 + 20 \cdot 2x + 15 \cdot 2x + 2x \cdot 2x
\]
\[
= 300 + 40x + 30x + 4x^2
\]
\[
= 4x^2 + 70x + 300
\]

So the equation becomes:
\[
4x^2 + 70x + 300 = 600
\]

#### Step 5: Simplify the equation
Subtract 600 from both sides to set the equation to zero:
\[
4x^2 + 70x + 300 - 600 = 0
\]
\[
4x^2 + 70x - 300 = 0
\]

#### Step 6: Solve the quadratic equation
The quadratic equation is:
\[
4x^2 + 70x - 300 = 0
\]
To simplify, divide the entire equation by 2:
\[
2x^2 + 35x - 150 = 0
\]

We solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = 35 \), and \( c = -150 \).

First, calculate the discriminant:
\[
b^2 - 4ac = 35^2 - 4 \cdot 2 \cdot (-150)
\]
\[
= 1225 + 1200
\]
\[
= 2425
\]

Next, find the square root of the discriminant:
\[
\sqrt{2425} = 5\sqrt{97}
\]

Now, apply the quadratic formula:
\[
x = \frac{-35 \pm 5\sqrt{97}}{2 \cdot 2}
\]
\[
x = \frac{-35 \pm 5\sqrt{97}}{4}
\]

This gives two potential solutions:
\[
x = \frac{-35 + 5\sqrt{97}}{4} \quad \text{and} \quad x = \frac{-35 - 5\sqrt{97}}{4}
\]

Since \( x \) represents the width of the path, it must be a positive value. Therefore, we discard the negative solution:
\[
x = \frac{-35 + 5\sqrt{97}}{4}
\]

#### Step 7: Approximate the value of \( x \)
To find an approximate numerical value, we calculate:
\[
\sqrt{97} \approx 9.849
\]
\[
-35 + 5\sqrt{97} \approx -35 + 5 \cdot 9.849 = -35 + 49.245 = 14.245
\]
\[
x \approx \frac{14.245}{4} \approx 3.561
\]

Rounding to a reasonable precision, we get:
\[
x \approx 3.56 \text{ meters}
\]

#### Final Answer:
\[
\boxed{3}
\]

(Note: The exact value is \( x = \frac{-35 + 5\sqrt{97}}{4} \), but for practical purposes, the width of the path is approximately 3 meters.)