Form the quadratic equation.

The area of a rectangle is given by the formula $A = L \times W$.
Given:
Length $L = (x+4)$ m
Width $W = (x-3)$ m
Area $A = 77 \text{ m}^2$

Substitute these values into the area formula:

$$(x+4)(x-3) = 77$$

Expand the left side of the equation:

$$x^2 - 3x + 4x - 12 = 77$$ $$x^2 + x - 12 = 77$$

Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$:

$$x^2 + x - 12 - 77 = 0$$ $$x^2 + x - 89 = 0$$

The quadratic equation is $x^2 + x - 89 = 0$.
We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=1$, $b=1$, and $c=-89$.

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-89)}}{2(1)}$$ $$x = \frac{-1 \pm \sqrt{1 + 356}}{2}$$ $$x = \frac{-1 \pm \sqrt{357}}{2}$$

Calculate the value of $\sqrt{357}$:

$$\sqrt{357} \approx 18.894$$

Now find the two possible values for $x$:

$$x_1 = \frac{-1 + 18.894}{2} = \frac{17.894}{2} = 8.947$$ $$x_2 = \frac{-1 - 18.894}{2} = \frac{-19.894}{2} = -9.947$$

Since $x$ represents a dimension, it must be positive. Therefore, $x \approx 8.947$.

Length $L = x+4$

$$L = 8.947 + 4 = 12.947 \text{ m}$$

Width $W = x-3$

$$W = 8.947 - 3 = 5.947 \text{ m}$$

The dimensions of the hall are approximately:
Length: $\boxed{\text{12.95 m}}$
Width: $\boxed{\text{5.95 m}}$