Complete the Square to Solve Equations worksheet featuring 10 quadratic equations.

To solve the given quadratic equations using the method of completing the square, we will follow a systematic approach for each equation. Let's go through each problem step by step.

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General Steps for Completing the Square:

1. Ensure the equation is in the form \( y^2 + by + c = 0 \).
2. Move the constant term \( c \) to the right side of the equation.
3. Add and subtract \( \left(\frac{b}{2}\right)^2 \) on the left side to complete the square.
4. Rewrite the left side as a perfect square trinomial.
5. Solve for \( y \) using the square root property.

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Problem 1: \( y^2 + 12y - 64 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 + 12y = 64 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is 12. Half of 12 is 6, and squaring it gives \( 6^2 = 36 \). Add 36 to both sides:
\[ y^2 + 12y + 36 = 64 + 36 \]
\[ y^2 + 12y + 36 = 100 \]

#### Step 3: Write as a perfect square.
\[ (y + 6)^2 = 100 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y + 6 = \pm \sqrt{100} \]
\[ y + 6 = \pm 10 \]

Solve for \( y \):
\[ y = -6 + 10 \quad \text{or} \quad y = -6 - 10 \]
\[ y = 4 \quad \text{or} \quad y = -16 \]

#### Solution:
\[ \boxed{4, -16} \]

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Problem 2: \( y^2 - 14y + 40 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 - 14y = -40 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is -14. Half of -14 is -7, and squaring it gives \( (-7)^2 = 49 \). Add 49 to both sides:
\[ y^2 - 14y + 49 = -40 + 49 \]
\[ y^2 - 14y + 49 = 9 \]

#### Step 3: Write as a perfect square.
\[ (y - 7)^2 = 9 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y - 7 = \pm \sqrt{9} \]
\[ y - 7 = \pm 3 \]

Solve for \( y \):
\[ y = 7 + 3 \quad \text{or} \quad y = 7 - 3 \]
\[ y = 10 \quad \text{or} \quad y = 4 \]

#### Solution:
\[ \boxed{10, 4} \]

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Problem 3: \( y^2 - 14y + 40 = 0 \)

This is the same as Problem 2. The solution is:
\[ \boxed{10, 4} \]

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Problem 4: \( y^2 + 8y + 15 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 + 8y = -15 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is 8. Half of 8 is 4, and squaring it gives \( 4^2 = 16 \). Add 16 to both sides:
\[ y^2 + 8y + 16 = -15 + 16 \]
\[ y^2 + 8y + 16 = 1 \]

#### Step 3: Write as a perfect square.
\[ (y + 4)^2 = 1 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y + 4 = \pm \sqrt{1} \]
\[ y + 4 = \pm 1 \]

Solve for \( y \):
\[ y = -4 + 1 \quad \text{or} \quad y = -4 - 1 \]
\[ y = -3 \quad \text{or} \quad y = -5 \]

#### Solution:
\[ \boxed{-3, -5} \]

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Problem 5: \( y^2 - 2y - 15 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 - 2y = 15 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is -2. Half of -2 is -1, and squaring it gives \( (-1)^2 = 1 \). Add 1 to both sides:
\[ y^2 - 2y + 1 = 15 + 1 \]
\[ y^2 - 2y + 1 = 16 \]

#### Step 3: Write as a perfect square.
\[ (y - 1)^2 = 16 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y - 1 = \pm \sqrt{16} \]
\[ y - 1 = \pm 4 \]

Solve for \( y \):
\[ y = 1 + 4 \quad \text{or} \quad y = 1 - 4 \]
\[ y = 5 \quad \text{or} \quad y = -3 \]

#### Solution:
\[ \boxed{5, -3} \]

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Problem 6: \( y^2 + 6y - 7 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 + 6y = 7 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is 6. Half of 6 is 3, and squaring it gives \( 3^2 = 9 \). Add 9 to both sides:
\[ y^2 + 6y + 9 = 7 + 9 \]
\[ y^2 + 6y + 9 = 16 \]

#### Step 3: Write as a perfect square.
\[ (y + 3)^2 = 16 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y + 3 = \pm \sqrt{16} \]
\[ y + 3 = \pm 4 \]

Solve for \( y \):
\[ y = -3 + 4 \quad \text{or} \quad y = -3 - 4 \]
\[ y = 1 \quad \text{or} \quad y = -7 \]

#### Solution:
\[ \boxed{1, -7} \]

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Problem 7: \( y^2 - 16y + 48 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 - 16y = -48 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is -16. Half of -16 is -8, and squaring it gives \( (-8)^2 = 64 \). Add 64 to both sides:
\[ y^2 - 16y + 64 = -48 + 64 \]
\[ y^2 - 16y + 64 = 16 \]

#### Step 3: Write as a perfect square.
\[ (y - 8)^2 = 16 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y - 8 = \pm \sqrt{16} \]
\[ y - 8 = \pm 4 \]

Solve for \( y \):
\[ y = 8 + 4 \quad \text{or} \quad y = 8 - 4 \]
\[ y = 12 \quad \text{or} \quad y = 4 \]

#### Solution:
\[ \boxed{12, 4} \]

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Problem 8: \( y^2 - 12y - 28 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 - 12y = 28 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is -12. Half of -12 is -6, and squaring it gives \( (-6)^2 = 36 \). Add 36 to both sides:
\[ y^2 - 12y + 36 = 28 + 36 \]
\[ y^2 - 12y + 36 = 64 \]

#### Step 3: Write as a perfect square.
\[ (y - 6)^2 = 64 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y - 6 = \pm \sqrt{64} \]
\[ y - 6 = \pm 8 \]

Solve for \( y \):
\[ y = 6 + 8 \quad \text{or} \quad y = 6 - 8 \]
\[ y = 14 \quad \text{or} \quad y = -2 \]

#### Solution:
\[ \boxed{14, -2} \]

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Problem 9: \( y^2 - 14y + 13 = 0 \)

#### Step 1: Move the constant term to the right.
\[ y^2 - 14y = -13 \]

#### Step 2: Complete the square.
The coefficient of \( y \) is -14. Half of -14 is -7, and squaring it gives \( (-7)^2 = 49 \). Add 49 to both sides:
\[ y^2 - 14y + 49 = -13 + 49 \]
\[ y^2 - 14y + 49 = 36 \]

#### Step 3: Write as a perfect square.
\[ (y - 7)^2 = 36 \]

#### Step 4: Solve for \( y \).
Take the square root of both sides:
\[ y - 7 = \pm \sqrt{36} \]
\[ y - 7 = \pm 6 \]

Solve for \( y \):
\[ y = 7 + 6 \quad \text{or} \quad y = 7 - 6 \]
\[ y = 13 \quad \text{or} \quad y = 1 \]

#### Solution:
\[ \boxed{13, 1} \]

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Problem 10: \( y^2 - 12y - 28 = 0 \)

This is the same as Problem 8. The solution is:
\[ \boxed{14, -2} \]

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Final Answers:

1. \( \boxed{4, -16} \)
2. \( \boxed{10, 4} \)
3. \( \boxed{10, 4} \)
4. \( \boxed{-3, -5} \)
5. \( \boxed{5, -3} \)
6. \( \boxed{1, -7} \)
7. \( \boxed{12, 4} \)
8. \( \boxed{14, -2} \)
9. \( \boxed{13, 1} \)
10. \( \boxed{14, -2} \)