In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)-14x+48=0`
(ii) `y^(2)+6=5y`

Answer

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

### Step 1: Solve the first equation \( x^2 - 14x + 48 = 0 \)

1. **Identify the equation**: The equation is \( x^2 - 14x + 48 = 0 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 48 \) (the constant term) and add up to \( -14 \) (the coefficient of \( x \)). The numbers are \( -6 \) and \( -8 \).
3. **Write the factored form**: The equation can be factored as:
\[ (x - 6)(x - 8) = 0 \]
4. **Set each factor to zero**:
- \( x - 6 = 0 \) gives \( x = 6 \)
- \( x - 8 = 0 \) gives \( x = 8 \)

Thus, the solutions for the first equation are:
\[ x = 6 \quad \text{and} \quad x = 8 \]

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

1. **Rearrange the equation**: Start by moving all terms to one side:
\[ y^2 - 5y + 6 = 0 \]
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 6 \) and add up to \( -5 \). The numbers are \( -2 \) and \( -3 \).
3. **Write the factored form**: The equation can be factored as:
\[ (y - 2)(y - 3) = 0 \]
4. **Set each factor to zero**:
- \( y - 2 = 0 \) gives \( y = 2 \)
- \( y - 3 = 0 \) gives \( y = 3 \)

Thus, the solutions for the second equation are:
\[ y = 2 \quad \text{and} \quad y = 3 \]

### Final Solutions
- For the first equation, the solutions are \( x = 6 \) and \( x = 8 \).
- For the second equation, the solutions are \( y = 2 \) and \( y = 3 \).

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