To solve this problem, let's analyze the provided information step-by-step.

Step 1: Define Variables

1. Let $w$ be the width of the box (given in inches).
2. Length is defined as $w + 20$. So, the length is 20 inches more than the width.
3. Height is given as 12 inches.

Step 2: Calculate the Volume

The formula for the volume of a rectangular box is: $\text{Volume} = \text{Width} \times \text{Length} \times \text{Height}$

Plugging in the values: $\text{Volume} = w \times (w + 20) \times 12$

Step 3: Set Up the Equation

The problem likely provides a target volume that the box needs to reach. From the image, we can see an equation: $w^2 + 20w - 300 = 0$

This equation results from setting the volume expression to a specific value (likely 600 cubic inches) and simplifying.

1. Substitute the volume expression into an equation: $w \times (w + 20) \times 12 = 600$

2. Divide both sides by 12 to simplify: $w \times (w + 20) = 50$

3. Expand and rearrange to form a quadratic equation: $w^2 + 20w - 300 = 0$

Step 4: Solve the Quadratic Equation

We have a quadratic equation in the form $w^2 + 20w - 300 = 0$. Solving this using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 20$, and $c = -300$.

1. Calculate the discriminant: $b^2 - 4ac = 20^2 - 4 \times 1 \times (-300) = 400 + 1200 = 1600$

2. Take the square root of the discriminant: $\sqrt{1600} = 40$

3. Substitute into the quadratic formula: $w = \frac{-20 \pm 40}{2}$

4. This gives two solutions: $w = \frac{20}{2} = 10 \quad \text{and} \quad w = \frac{-60}{2} = -30$

Since width cannot be negative, we discard $w = -30$, leaving: $w = 10$

Conclusion

The width of the box, $w$, is 10 inches. The length is $w + 20 = 30$ inches, and the height is 12 inches. These dimensions satisfy the volume requirement.

Would you like more details or have any questions?

Here are 5 related questions to expand on the topic:

1. How do you derive the quadratic formula from a general quadratic equation?
2. What are alternative methods to solve quadratic equations besides the quadratic formula?
3. How can we verify the solution by substituting back into the volume formula?
4. How do different values of $w$ affect the box’s volume?
5. What happens if we change the height or target volume?

Tip: When solving quadratic equations, always check if factoring is possible before using the quadratic formula for a quicker solution.