(i) `x^(2)-7x+12=0`
(ii) `y^(2)+y-12=0`

Options

A. `xgey`
B. `xgty`
C. `xley`
D. `xlty`

Answer

To solve the given equations step by step, we will tackle each equation separately.

### Step 1: Solve the first equation \( x^2 - 7x + 12 = 0 \)

1. **Identify the coefficients**:
The equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -7 \)
- \( c = 12 \)

2. **Factor the quadratic**:
We need to find two numbers that multiply to \( c \) (12) and add up to \( b \) (-7). The numbers are -3 and -4.
Thus, we can factor the equation as:
\[ (x - 3)(x - 4) = 0 \]

3. **Set each factor to zero**:
\[ x - 3 = 0 \quad \text{or} \quad x - 4 = 0 \]

4. **Solve for \( x \)**:
\[ x = 3 \quad \text{or} \quad x = 4 \]

### Step 2: Solve the second equation \( y^2 + y - 12 = 0 \)

1. **Identify the coefficients**:
The equation is in the standard form \( ay^2 + by + c = 0 \), where:
- \( a = 1 \)
- \( b = 1 \)
- \( c = -12 \)

2. **Factor the quadratic**:
We need to find two numbers that multiply to \( c \) (-12) and add up to \( b \) (1). The numbers are 4 and -3.
Thus, we can factor the equation as:
\[ (y + 4)(y - 3) = 0 \]

3. **Set each factor to zero**:
\[ y + 4 = 0 \quad \text{or} \quad y - 3 = 0 \]

4. **Solve for \( y \)**:
\[ y = -4 \quad \text{or} \quad y = 3 \]

### Summary of Solutions:
- For the first equation \( x^2 - 7x + 12 = 0 \), the solutions are \( x = 3 \) and \( x = 4 \).
- For the second equation \( y^2 + y - 12 = 0 \), the solutions are \( y = -4 \) and \( y = 3 \).