Walter is making an open-top box. The length is 6 inches greater than the width, and the height is 2 inches. The area of the rectangular sheet of material that will be folded to make the box is 320 square inches.

What is the width of the box?

To find the width of the box, we can start by setting up an equation based on the given information.

Let's assume the width of the box is "x" inches.

According to the problem, the length of the box is 6 inches greater than the width. So, the length would be (x + 6) inches.

The height of the box is given as 2 inches.

To find the area of the rectangular sheet, we need to calculate the product of the length and width. According to the problem, the area is 320 square inches.

So, the equation would be:
Length × Width = Area
(x + 6) × x = 320

Now, we can solve this equation to find the width of the box.

Expanding the equation, we have:
x^2 + 6x = 320

Rearranging the equation to make it equal to zero:
x^2 + 6x - 320 = 0

Now, we can either factorize the equation or use the quadratic formula to solve for x. Factoring might not be possible in this case, so let's use the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

For the equation x^2 + 6x - 320 = 0, a = 1, b = 6, and c = -320.

Plugging these values into the quadratic formula, we get:
x = [-6 ± √(6^2 - 4(1)(-320))] / 2(1)
x = [-6 ± √(36 + 1280)] / 2
x = [-6 ± √(1316)] / 2

Now, we can simplify further:
x = (-6 ± √(1316)) / 2
x = (-6 ± √(4 × 329)) / 2
x = (-6 ± 2√(329)) / 2
x = -3 ± √(329)

Since the width of a box cannot be negative, we discard the negative value.

Therefore, the width of the box is approximately x ≈ -3 + √(329), which is approximately 16.14 inches.

So, the width of the box is approximately 16.14 inches.