Step 1: 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 $ax^2 + bx + c = 0$: $$x^2 + x - 12 - 77 = 0$$ $$x^2 + x - 89 = 0$$ Step 2: Solve the quadratic equation for $x$. 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, $x \approx 8.947$. Step 3: Find the dimensions of the hall. 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: $\boxed{\text{5.95 m}}$ That's 2 down. 3 left today — send the next one.