Find the zeros of `2x^(2) - x - 45`.

Answer

To find the zeros of the quadratic equation \(2x^2 - x - 45\), we will follow these steps:

### Step 1: Identify coefficients
The given quadratic equation is in the standard form \(ax^2 + bx + c\), where:
- \(a = 2\)
- \(b = -1\)
- \(c = -45\)

### Step 2: Calculate \(a \cdot c\)
Next, we need to calculate the product of \(a\) and \(c\):
\[ a \cdot c = 2 \cdot (-45) = -90 \]

### Step 3: Find two numbers that multiply to \(a \cdot c\) and add to \(b\)
We need to find two numbers \(b_1\) and \(b_2\) such that:
- \(b_1 \cdot b_2 = -90\)
- \(b_1 + b_2 = -1\)

After testing pairs of factors of \(-90\), we find:
- \(b_1 = -10\)
- \(b_2 = 9\)

This is because:
\[ -10 \cdot 9 = -90 \quad \text{and} \quad -10 + 9 = -1 \]

### Step 4: Rewrite the middle term using \(b_1\) and \(b_2\)
Now we can rewrite the equation by replacing \(-x\) with \(-10x + 9x\):
\[ 2x^2 - 10x + 9x - 45 \]

### Step 5: Factor by grouping
Next, we will group the terms:
\[ (2x^2 - 10x) + (9x - 45) \]
Now, factor out the common terms from each group:
\[ 2x(x - 5) + 9(x - 5) \]

### Step 6: Factor out the common binomial
Now we can factor out the common binomial \((x - 5)\):
\[ (x - 5)(2x + 9) \]

### Step 7: Set each factor to zero
To find the zeros, we set each factor equal to zero:
1. \(x - 5 = 0\)
2. \(2x + 9 = 0\)

### Step 8: Solve for \(x\)
Solving these equations gives:
1. From \(x - 5 = 0\):
\[ x = 5 \]
2. From \(2x + 9 = 0\):
\[ 2x = -9 \implies x = -\frac{9}{2} \]

### Conclusion
The zeros of the equation \(2x^2 - x - 45\) are:
\[ x = 5 \quad \text{and} \quad x = -\frac{9}{2} \]