In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)=15x-56=0`
(ii) `y^(2)-23y+132=0`

Answer

To solve the given equations step by step, we will tackle each equation one at a time.

### Step 1: Solve Equation I
The first equation is:
\[ x^2 = 15x - 56 \]

Rearranging the equation, we get:
\[ x^2 - 15x + 56 = 0 \]

### Step 2: Factor the Quadratic
We need to factor the quadratic equation \( x^2 - 15x + 56 \). We are looking for two numbers that multiply to \( 56 \) and add up to \( 15 \).

The numbers \( 7 \) and \( 8 \) satisfy this condition:
\[ 7 \times 8 = 56 \]
\[ 7 + 8 = 15 \]

Thus, we can factor the equation as:
\[ (x - 7)(x - 8) = 0 \]

### Step 3: Find the Values of x
Setting each factor to zero gives us:
1. \( x - 7 = 0 \) → \( x = 7 \)
2. \( x - 8 = 0 \) → \( x = 8 \)

So, the solutions for the first equation are:
\[ x = 7 \quad \text{and} \quad x = 8 \]

### Step 4: Solve Equation II
The second equation is:
\[ y^2 - 23y + 132 = 0 \]

### Step 5: Factor the Quadratic
We need to factor the quadratic equation \( y^2 - 23y + 132 \). We are looking for two numbers that multiply to \( 132 \) and add up to \( 23 \).

The numbers \( 11 \) and \( 12 \) satisfy this condition:
\[ 11 \times 12 = 132 \]
\[ 11 + 12 = 23 \]

Thus, we can factor the equation as:
\[ (y - 11)(y - 12) = 0 \]

### Step 6: Find the Values of y
Setting each factor to zero gives us:
1. \( y - 11 = 0 \) → \( y = 11 \)
2. \( y - 12 = 0 \) → \( y = 12 \)

So, the solutions for the second equation are:
\[ y = 11 \quad \text{and} \quad y = 12 \]

### Final Solution
The solutions to the equations are:
- For Equation I: \( x = 7 \) and \( x = 8 \)
- For Equation II: \( y = 11 \) and \( y = 12 \)

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