A rectangular yard has a length of (x+2) m and a width of (x-3) m. If the area of the yard is 40 m², find the value of x.

The length ($L$) of the yard is $(x+2)$ m and the width ($W$) is $(x-3)$ m. The area ($A$) is given as $40 \text{ m}^2$.
The formula for the area of a rectangle is $A = L \times W$.
Substitute the given expressions into the formula:

$$(x+2)(x-3) = 40$$

Multiply the binomials:

$$x^2 - 3x + 2x - 6 = 40$$ $$x^2 - x - 6 = 40$$

Subtract 40 from both sides:

$$x^2 - x - 6 - 40 = 0$$ $$x^2 - x - 46 = 0$$

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
For $x^2 - x - 46 = 0$, we have $a=1$, $b=-1$, and $c=-46$.

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-46)}}{2(1)}$$ $$x = \frac{1 \pm \sqrt{1 + 184}}{2}$$ $$x = \frac{1 \pm \sqrt{185}}{2}$$

We have two possible values for $x$: $x_1 = \frac{1 + \sqrt{185}}{2}$ and $x_2 = \frac{1 - \sqrt{185}}{2}$.
Since $\sqrt{185} \approx 13.60$, let's evaluate these:
$x_1 \approx \frac{1 + 13.60}{2} = \frac{14.60}{2} = 7.30$
$x_2 \approx \frac{1 - 13.60}{2} = \frac{-12.60}{2} = -6.30$

The width of the yard is $(x-3)$ m. If we use $x = -6.30$, the width would be $-6.30 - 3 = -9.30$ m, which is not possible for a physical dimension.
Therefore, we must use the positive value for $x$:

$$x = \frac{1 + \sqrt{185}}{2}$$

Substitute the valid value of $x$ back into the expressions for length and width:
Length $L = x+2 = \left(\frac{1 + \sqrt{185}}{2}\right) + 2 = \frac{1 + \sqrt{185} + 4}{2} = \frac{5 + \sqrt{185}}{2}$ m
Width $W = x-3 = \left(\frac{1 + \sqrt{185}}{2}\right) - 3 = \frac{1 + \sqrt{185} - 6}{2} = \frac{-5 + \sqrt{185}}{2}$ m

The length and width of the yard are:
Length: $\boxed{\frac{5 + \sqrt{185}}{2} \text{ m}}$
Width: $\boxed{\frac{-5 + \sqrt{185}}{2} \text{ m}}$