Let the width of the garden be denoted by $w$ feet. According to the problem, the length of the garden is 2 feet more than the width, so the length is $w + 2$ feet.

The area of a rectangle is given by:

$\text{Area} = \text{length} \times \text{width}$

Substituting the given values:

$288 = w \times (w + 2)$

This expands to the quadratic equation:

$288 = w^2 + 2w$

Rearranging the equation:

$w^2 + 2w - 288 = 0$

Now, solve this quadratic equation using the quadratic formula:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = 2$, and $c = -288$. Substituting these values:

$w = \frac{-2 \pm \sqrt{2^2 - 4(1)(-288)}}{2(1)}$ $w = \frac{-2 \pm \sqrt{4 + 1152}}{2}$ $w = \frac{-2 \pm \sqrt{1156}}{2}$ $w = \frac{-2 \pm 34}{2}$

Thus, the two possible solutions for $w$ are:

$w = \frac{-2 + 34}{2} = \frac{32}{2} = 16$ or $w = \frac{-2 - 34}{2} = \frac{-36}{2} = -18$

Since the width cannot be negative, we have $w = 16$ feet.

Therefore, the width is 16 feet, and the length is:

$\text{length} = w + 2 = 16 + 2 = 18 \, \text{feet}.$

Thus, the dimensions of the garden are:

$\text{Length} = 18 \, \text{feet}, \quad \text{Width} = 16 \, \text{feet}.$

Would you like more details or have any questions?

Related questions:

1. How would the problem change if the length were 3 feet more than the width?
2. How can you derive the quadratic equation from the original area equation?
3. What are the methods for solving quadratic equations besides the quadratic formula?
4. What would the dimensions be if the area were doubled?
5. How would you calculate the perimeter of the garden?

Tip:

When solving quadratic equations, always check the physical context (e.g., a negative length doesn't make sense in real-world problems).