In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)+12x=-36`
(ii) `y^(2)+15y+56=0`

Answer

To solve the given equations step by step, we will start with the first equation and then proceed to the second one.

### Step 1: Solve the first equation \( x^2 + 12x = -36 \)

1. **Rearrange the equation**:
We want to bring all terms to one side of the equation.
\[ x^2 + 12x + 36 = 0 \]

2. **Identify the coefficients**:
Here, we have a quadratic equation in the form \( ax^2 + bx + c = 0 \) where \( a = 1 \), \( b = 12 \), and \( c = 36 \).

3. **Factor the quadratic**:
We need to find two numbers that multiply to \( c \) (36) and add up to \( b \) (12). The numbers are 6 and 6.
\[ (x + 6)(x + 6) = 0 \]

4. **Set each factor to zero**:
Since both factors are the same, we have:
\[ x + 6 = 0 \]

5. **Solve for \( x \)**:
\[ x = -6 \]

### Step 2: Solve the second equation \( y^2 + 15y + 56 = 0 \)

1. **Identify the coefficients**:
This is also a quadratic equation where \( a = 1 \), \( b = 15 \), and \( c = 56 \).

2. **Factor the quadratic**:
We need to find two numbers that multiply to \( c \) (56) and add up to \( b \) (15). The numbers are 8 and 7.
\[ (y + 8)(y + 7) = 0 \]

3. **Set each factor to zero**:
\[ y + 8 = 0 \quad \text{or} \quad y + 7 = 0 \]

4. **Solve for \( y \)**:
\[ y = -8 \quad \text{or} \quad y = -7 \]

### Final Answers:
- For the first equation, \( x = -6 \).
- For the second equation, \( y = -8 \) and \( y = -7 \).