What are the roots of the equation `2(y+2)^(2) - 5 (y+2) = 12 ` ?

Options

A. `-7//2,2`
B. `-3//2,4`
C. `-5//3,3`
D. `3//2,4`

Explanation

To find the roots of the equation \(2(y+2)^{2} - 5(y+2) = 12\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2(y+2)^{2} - 5(y+2) = 12 \]

### Step 2: Move all terms to one side
We can rearrange the equation to set it to zero:
\[ 2(y+2)^{2} - 5(y+2) - 12 = 0 \]

### Step 3: Substitute to simplify
Let \(x = y + 2\). Then the equation becomes:
\[ 2x^{2} - 5x - 12 = 0 \]

### Step 4: Factor the quadratic equation
Next, we need to factor the quadratic equation \(2x^{2} - 5x - 12\). We look for two numbers that multiply to \(2 \times -12 = -24\) and add to \(-5\). The numbers \(-8\) and \(3\) work:
\[ 2x^{2} - 8x + 3x - 12 = 0 \]
Now, we can group the terms:
\[ (2x^{2} - 8x) + (3x - 12) = 0 \]
Factoring by grouping gives:
\[ 2x(x - 4) + 3(x - 4) = 0 \]
Factoring out the common term \((x - 4)\):
\[ (2x + 3)(x - 4) = 0 \]

### Step 5: Solve for \(x\)
Setting each factor to zero gives us:
1. \(2x + 3 = 0\)
\[ 2x = -3 \implies x = -\frac{3}{2} \]
2. \(x - 4 = 0\)
\[ x = 4 \]

### Step 6: Substitute back to find \(y\)
Now, we substitute back to find \(y\):
1. For \(x = -\frac{3}{2}\):
\[ y + 2 = -\frac{3}{2} \implies y = -\frac{3}{2} - 2 = -\frac{3}{2} - \frac{4}{2} = -\frac{7}{2} \]
2. For \(x = 4\):
\[ y + 2 = 4 \implies y = 4 - 2 = 2 \]

### Conclusion
The roots of the equation are:
\[ y = -\frac{7}{2} \quad \text{and} \quad y = 2 \]