In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)+7x+12=0`
(ii) `y^(2)+6y+8=0`

Answer

To solve the given equations step by step, we will follow the method of factoring for both quadratic equations.

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

1. **Identify the coefficients**: In the equation \( x^2 + 7x + 12 = 0 \), we have:
- Coefficient of \( x^2 \) (a) = 1
- Coefficient of \( x \) (b) = 7
- Constant term (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 because:
- \( 3 \times 4 = 12 \)
- \( 3 + 4 = 7 \)

3. **Rewrite the equation**: We can rewrite the equation as:
\[ (x + 3)(x + 4) = 0 \]

4. **Set each factor to zero**:
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x + 4 = 0 \) → \( x = -4 \)

So, the solutions for the first equation are:
\[ x = -3 \quad \text{and} \quad x = -4 \]

### Step 2: Solve the second equation \( y^2 + 6y + 8 = 0 \)

1. **Identify the coefficients**: In the equation \( y^2 + 6y + 8 = 0 \), we have:
- Coefficient of \( y^2 \) (a) = 1
- Coefficient of \( y \) (b) = 6
- Constant term (c) = 8

2. **Factor the quadratic**: We need to find two numbers that multiply to \( c \) (8) and add up to \( b \) (6). The numbers are 2 and 4 because:
- \( 2 \times 4 = 8 \)
- \( 2 + 4 = 6 \)

3. **Rewrite the equation**: We can rewrite the equation as:
\[ (y + 2)(y + 4) = 0 \]

4. **Set each factor to zero**:
- \( y + 2 = 0 \) → \( y = -2 \)
- \( y + 4 = 0 \) → \( y = -4 \)

So, the solutions for the second equation are:
\[ y = -2 \quad \text{and} \quad y = -4 \]

### Final Solutions:
- For the first equation \( x^2 + 7x + 12 = 0 \): \( x = -3 \) and \( x = -4 \)
- For the second equation \( y^2 + 6y + 8 = 0 \): \( y = -2 \) and \( y = -4 \)

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