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The problem provides the equation for monthly profit, $ P(n) = -0.01n^2 + 8n - 700 $, where $ n $ is the number of items put up for sale, and $ P(n) $ is the profit in dollars. It asks us to find $ n $ when the profit is $500. 

### Step 1: Set the equation equal to 500.
\[
500 = -0.01n^2 + 8n - 700
\]

### Step 2: Rearrange the equation into standard quadratic form.
\[
-0.01n^2 + 8n - 700 - 500 = 0
\]
\[
-0.01n^2 + 8n - 1200 = 0
\]

### Step 3: Multiply through by -100 to eliminate the decimals.
\[
n^2 - 800n + 120,000 = 0
\]

### Step 4: Solve the quadratic equation using the quadratic formula, $ n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $, where $ A = 1 $, $ B = -800 $, and $ C = 120,000 $.

\[
n = \frac{800 \pm \sqrt{(-800)^2 - 4 \cdot 1 \cdot 120,000}}{2 \cdot 1}
\]
\[
n = \frac{800 \pm \sqrt{640,000 - 480,000}}{2}
\]
\[
n = \frac{800 \pm \sqrt{160,000}}{2}
\]
\[
n = \frac{800 \pm 400}{2}
\]

### Step 5: Calculate the two possible values of $ n $.
1. $ n = \frac{800 + 400}{2} = \frac{1200}{2} = 600 $
2. $ n = \frac{800 - 400}{2} = \frac{400}{2} = 200 $

### Final Answer
The number of items put up for sale could be **200 or 600 items**.

So, the correct answer is:

**(e) 200 or 600 items**

Would you like a detailed explanation of any step? 

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Here are five related questions to consider:
1. How do you determine the maximum or minimum profit from a quadratic function?
2. What is the significance of the discriminant in a quadratic equation?
3. How can you use graphing to solve quadratic equations?
4. Why does this problem have two valid solutions for $ n $?
5. How would the profit change if the cost per item increased?

**Tip**: When solving a quadratic equation, always check if factoring is possible before using the quadratic formula—it can save time!

The equation P(n) = -0.01n^2 + 8n - 700 takes n, the number of items that a retailer puts up for sale in their online store each month, and gives P, the store's monthly profit (in dollars). If the store generated $500 in profit last month, how many items did they put up for sale? Round your answer to the nearest whole item. If there is more than one possible answer, give both.

Learn how to solve for n in the quadratic profit equation P(n) = -0.01n^2 + 8n - 700, where n represents the number of items sold and P(n) is the profit. Given a profit of $500, find how many items were put up for sale. Suitable for Grades 9-11.

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