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)-12y+32=0`

Answer

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

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

1. **Identify the equation**:
\[ x^2 - 7x + 12 = 0 \]

2. **Factor the quadratic**:
We need two numbers that multiply to \( 12 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)). The numbers \( -3 \) and \( -4 \) satisfy these conditions:
\[ (x - 3)(x - 4) = 0 \]

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 \( y^2 - 12y + 32 = 0 \)

1. **Identify the equation**:
\[ y^2 - 12y + 32 = 0 \]

2. **Factor the quadratic**:
We need two numbers that multiply to \( 32 \) and add up to \( -12 \). The numbers \( -4 \) and \( -8 \) satisfy these conditions:
\[ (y - 4)(y - 8) = 0 \]

3. **Set each factor to zero**:
\[ y - 4 = 0 \quad \text{or} \quad y - 8 = 0 \]

4. **Solve for \( y \)**:
\[ y = 4 \quad \text{or} \quad y = 8 \]

### Final Solutions
- For the first equation, the solutions are \( x = 3 \) and \( x = 4 \).
- For the second equation, the solutions are \( y = 4 \) and \( y = 8 \).