In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)-x-12=0`
(ii) `y^(2)+5y+6=0`

Answer

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

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

1. **Factor the quadratic equation**: We need to find two numbers that multiply to \(-12\) (the constant term) and add to \(-1\) (the coefficient of \(x\)). The numbers that satisfy this are \(4\) and \(-3\).

\[ x^2 - x - 12 = (x - 4)(x + 3) = 0 \]

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

Thus, the solutions for \(x\) are:
\[ x = 4 \quad \text{or} \quad x = -3 \]

### Step 2: Solve the second equation \( y^2 + 5y + 6 = 0 \)

1. **Factor the quadratic equation**: We need to find two numbers that multiply to \(6\) (the constant term) and add to \(5\) (the coefficient of \(y\)). The numbers that satisfy this are \(3\) and \(2\).

\[ y^2 + 5y + 6 = (y + 3)(y + 2) = 0 \]

2. **Set each factor to zero**:
- \(y + 3 = 0 \Rightarrow y = -3\)
- \(y + 2 = 0 \Rightarrow y = -2\)

Thus, the solutions for \(y\) are:
\[ y = -3 \quad \text{or} \quad y = -2 \]

### Final Solutions

The solutions for the equations are:
- For \(x\): \(4\) or \(-3\)
- For \(y\): \(-3\) or \(-2\)