Find the zeros of `y=2x^(2)+x-6`.

Options

A. 3 and 2
B. `-3 and 2`
C. `(1)/(2) and (3)/(2)`
D. `(3)/(2) and -2`

Explanation

To find the zeros of the quadratic function \( y = 2x^2 + x - 6 \), we will follow these steps:

### Step 1: Identify the coefficients
The quadratic function is in the standard form \( ax^2 + bx + c \). Here, we have:
- \( a = 2 \)
- \( b = 1 \)
- \( c = -6 \)

### Step 2: Calculate \( a \cdot c \)
We need to calculate the product of \( a \) and \( c \):
\[ a \cdot c = 2 \cdot (-6) = -12 \]

### Step 3: Find two numbers that satisfy the conditions
We need to find two numbers \( b_1 \) and \( b_2 \) such that:
- \( b_1 + b_2 = b = 1 \)
- \( b_1 \cdot b_2 = a \cdot c = -12 \)

After inspecting the pairs of factors of -12, we find:
- \( b_1 = 4 \)
- \( b_2 = -3 \)

These satisfy:
\[ 4 + (-3) = 1 \quad \text{and} \quad 4 \cdot (-3) = -12 \]

### Step 4: Rewrite the quadratic equation
Now, we can rewrite the quadratic equation by breaking the middle term using \( b_1 \) and \( b_2 \):
\[ y = 2x^2 + 4x - 3x - 6 \]

### Step 5: Factor by grouping
Next, we will group the terms:
\[ y = (2x^2 + 4x) + (-3x - 6) \]
Now, factor out the common terms in each group:
\[ y = 2x(x + 2) - 3(x + 2) \]
Now, factor out \( (x + 2) \):
\[ y = (x + 2)(2x - 3) \]

### Step 6: Set the equation to zero
To find the zeros, we set \( y \) to zero:
\[ (x + 2)(2x - 3) = 0 \]

### Step 7: Solve for \( x \)
Now, we solve for \( x \) by setting each factor to zero:
1. \( x + 2 = 0 \) leads to \( x = -2 \)
2. \( 2x - 3 = 0 \) leads to \( 2x = 3 \) or \( x = \frac{3}{2} \)

### Final Answer
The zeros of the function \( y = 2x^2 + x - 6 \) are:
\[ x = -2 \quad \text{and} \quad x = \frac{3}{2} \]