Complete the square to solve quadratic equations worksheet.

Let's solve each of these quadratic equations by completing the square. The general method is:

Given:
$$ y^2 + by + c = 0 $$

1. Move constant to the other side:
$ y^2 + by = -c $

2. Take half of the coefficient of $ y $, square it, and add to both sides:
$ \left(\frac{b}{2}\right)^2 $

3. Factor the left side into a perfect square trinomial.

4. Solve for $ y $ using square roots.

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Let’s go through each problem step-by-step.

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1. $ y^2 + 12y - 64 = 0 $

Move constant:
$$
y^2 + 12y = 64
$$

Half of 12 is 6; square it: $ 6^2 = 36 $

Add 36 to both sides:
$$
y^2 + 12y + 36 = 64 + 36 = 100
$$

Factor:
$$
(y + 6)^2 = 100
$$

Take square root:
$$
y + 6 = \pm 10
\Rightarrow y = -6 \pm 10
$$

Solutions:
$$
y = 4 \quad \text{or} \quad y = -16
$$

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2. $ y^2 - 14y + 40 = 0 $

Move constant:
$$
y^2 - 14y = -40
$$

Half of -14 is -7; square: $ (-7)^2 = 49 $

Add 49:
$$
y^2 - 14y + 49 = -40 + 49 = 9
$$

Factor:
$$
(y - 7)^2 = 9
$$

Take square root:
$$
y - 7 = \pm 3
\Rightarrow y = 7 \pm 3
$$

Solutions:
$$
y = 10 \quad \text{or} \quad y = 4
$$

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3. $ y^2 - 14y + 40 = 0 $

This is the same as #2! So the answer is:

$$
y = 10 \quad \text{or} \quad y = 4
$$

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4. $ y^2 + 8y + 15 = 0 $

Move constant:
$$
y^2 + 8y = -15
$$

Half of 8 is 4; square: $ 4^2 = 16 $

Add 16:
$$
y^2 + 8y + 16 = -15 + 16 = 1
$$

Factor:
$$
(y + 4)^2 = 1
$$

Take square root:
$$
y + 4 = \pm 1
\Rightarrow y = -4 \pm 1
$$

Solutions:
$$
y = -3 \quad \text{or} \quad y = -5
$$

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5. $ y^2 - 2y - 15 = 0 $

Move constant:
$$
y^2 - 2y = 15
$$

Half of -2 is -1; square: $ (-1)^2 = 1 $

Add 1:
$$
y^2 - 2y + 1 = 15 + 1 = 16
$$

Factor:
$$
(y - 1)^2 = 16
$$

Take square root:
$$
y - 1 = \pm 4
\Rightarrow y = 1 \pm 4
$$

Solutions:
$$
y = 5 \quad \text{or} \quad y = -3
$$

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6. $ y^2 + 6y - 7 = 0 $

Move constant:
$$
y^2 + 6y = 7
$$

Half of 6 is 3; square: $ 3^2 = 9 $

Add 9:
$$
y^2 + 6y + 9 = 7 + 9 = 16
$$

Factor:
$$
(y + 3)^2 = 16
$$

Take square root:
$$
y + 3 = \pm 4
\Rightarrow y = -3 \pm 4
$$

Solutions:
$$
y = 1 \quad \text{or} \quad y = -7
$$

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7. $ y^2 - 16y + 48 = 0 $

Move constant:
$$
y^2 - 16y = -48
$$

Half of -16 is -8; square: $ (-8)^2 = 64 $

Add 64:
$$
y^2 - 16y + 64 = -48 + 64 = 16
$$

Factor:
$$
(y - 8)^2 = 16
$$

Take square root:
$$
y - 8 = \pm 4
\Rightarrow y = 8 \pm 4
$$

Solutions:
$$
y = 12 \quad \text{or} \quad y = 4
$$

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8. $ y^2 - 12y - 28 = 0 $

Move constant:
$$
y^2 - 12y = 28
$$

Half of -12 is -6; square: $ (-6)^2 = 36 $

Add 36:
$$
y^2 - 12y + 36 = 28 + 36 = 64
$$

Factor:
$$
(y - 6)^2 = 64
$$

Take square root:
$$
y - 6 = \pm 8
\Rightarrow y = 6 \pm 8
$$

Solutions:
$$
y = 14 \quad \text{or} \quad y = -2
$$

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9. $ y^2 - 14y + 13 = 0 $

Move constant:
$$
y^2 - 14y = -13
$$

Half of -14 is -7; square: $ (-7)^2 = 49 $

Add 49:
$$
y^2 - 14y + 49 = -13 + 49 = 36
$$

Factor:
$$
(y - 7)^2 = 36
$$

Take square root:
$$
y - 7 = \pm 6
\Rightarrow y = 7 \pm 6
$$

Solutions:
$$
y = 13 \quad \text{or} \quad y = 1
$$

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10. $ y^2 - 12y - 28 = 0 $

Same as #8!

So solutions:
$$
y = 14 \quad \text{or} \quad y = -2
$$

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✔ Final Answers:

| Problem | Solutions |
|--------|-----------|
| 1 | $ y = 4, -16 $ |
| 2 | $ y = 10, 4 $ |
| 3 | $ y = 10, 4 $ |
| 4 | $ y = -3, -5 $ |
| 5 | $ y = 5, -3 $ |
| 6 | $ y = 1, -7 $ |
| 7 | $ y = 12, 4 $ |
| 8 | $ y = 14, -2 $ |
| 9 | $ y = 13, 1 $ |
| 10 | $ y = 14, -2 $ |

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Let me know if you'd like these checked via factoring or quadratic formula for verification!