`x^(2)-2x-15=0`
`{:("Quantity A","Quantity B"),(x,1):}`

Answer

To solve the quadratic equation \(x^2 - 2x - 15 = 0\) and compare the values of \(x\) with \(1\), we will follow these steps:

### Step 1: Write the equation
We start with the equation:
\[ x^2 - 2x - 15 = 0 \]

### Step 2: Factor the quadratic equation
We need to factor the quadratic expression. We look for two numbers that multiply to \(-15\) (the constant term) and add to \(-2\) (the coefficient of \(x\)). The numbers that satisfy this are \(-5\) and \(3\). Thus, we can factor the equation as:
\[ (x - 5)(x + 3) = 0 \]

### Step 3: Set each factor to zero
Now we set each factor equal to zero:
1. \(x - 5 = 0\)
2. \(x + 3 = 0\)

### Step 4: Solve for \(x\)
Solving these equations gives us:
1. From \(x - 5 = 0\), we find \(x = 5\).
2. From \(x + 3 = 0\), we find \(x = -3\).

### Step 5: Compare the values of \(x\) with \(1\)
Now we have two possible values for \(x\):
- When \(x = 5\), \(x\) is greater than \(1\).
- When \(x = -3\), \(x\) is less than \(1\).

### Conclusion
Since we have two different values for \(x\) that yield different comparisons with \(1\), we cannot definitively say whether Quantity A (which is \(x\)) is greater than, less than, or equal to Quantity B (which is \(1\)). Therefore, the answer is that we cannot decide which quantity is greater.