Solve each quadratic equation by completing the square.

To solve each quadratic equation by completing the square, we follow a systematic approach. Here's how to solve each equation step by step:

---

1) \( k^2 - 2k - 8 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ k^2 - 2k = 8 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( k \), which is \(-2\), divide it by 2, and square it:
\[
\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1
\]
- Add this square to both sides:
\[
k^2 - 2k + 1 = 8 + 1
\]
\[
k^2 - 2k + 1 = 9
\]

#### Step 3: Write the left side as a perfect square.
\[ (k - 1)^2 = 9 \]

#### Step 4: Solve for \( k \).
- Take the square root of both sides:
\[
k - 1 = \pm \sqrt{9}
\]
\[
k - 1 = \pm 3
\]
- Solve for \( k \):
\[
k = 1 + 3 \quad \text{or} \quad k = 1 - 3
\]
\[
k = 4 \quad \text{or} \quad k = -2
\]

#### Final Answer:
\[
\boxed{k = 4 \text{ or } k = -2}
\]

---

2) \( x^2 - 2x = -26 \)

#### Step 1: Move the constant term to the right side (already done).
\[ x^2 - 2x = -26 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( x \), which is \(-2\), divide it by 2, and square it:
\[
\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1
\]
- Add this square to both sides:
\[
x^2 - 2x + 1 = -26 + 1
\]
\[
x^2 - 2x + 1 = -25
\]

#### Step 3: Write the left side as a perfect square.
\[ (x - 1)^2 = -25 \]

#### Step 4: Solve for \( x \).
- Take the square root of both sides:
\[
x - 1 = \pm \sqrt{-25}
\]
\[
x - 1 = \pm 5i
\]
- Solve for \( x \):
\[
x = 1 + 5i \quad \text{or} \quad x = 1 - 5i
\]

#### Final Answer:
\[
\boxed{x = 1 + 5i \text{ or } x = 1 - 5i}
\]

---

3) \( p^2 - p = 2 \)

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

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( p \), which is \(-1\), divide it by 2, and square it:
\[
\left(\frac{-1}{2}\right)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}
\]
- Add this square to both sides:
\[
p^2 - p + \frac{1}{4} = 2 + \frac{1}{4}
\]
\[
p^2 - p + \frac{1}{4} = \frac{8}{4} + \frac{1}{4}
\]
\[
p^2 - p + \frac{1}{4} = \frac{9}{4}
\]

#### Step 3: Write the left side as a perfect square.
\[ \left(p - \frac{1}{2}\right)^2 = \frac{9}{4} \]

#### Step 4: Solve for \( p \).
- Take the square root of both sides:
\[
p - \frac{1}{2} = \pm \sqrt{\frac{9}{4}}
\]
\[
p - \frac{1}{2} = \pm \frac{3}{2}
\]
- Solve for \( p \):
\[
p = \frac{1}{2} + \frac{3}{2} \quad \text{or} \quad p = \frac{1}{2} - \frac{3}{2}
\]
\[
p = 2 \quad \text{or} \quad p = -1
\]

#### Final Answer:
\[
\boxed{p = 2 \text{ or } p = -1}
\]

---

4) \( n^2 + n - 30 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ n^2 + n = 30 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( n \), which is \(1\), divide it by 2, and square it:
\[
\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}
\]
- Add this square to both sides:
\[
n^2 + n + \frac{1}{4} = 30 + \frac{1}{4}
\]
\[
n^2 + n + \frac{1}{4} = \frac{120}{4} + \frac{1}{4}
\]
\[
n^2 + n + \frac{1}{4} = \frac{121}{4}
\]

#### Step 3: Write the left side as a perfect square.
\[ \left(n + \frac{1}{2}\right)^2 = \frac{121}{4} \]

#### Step 4: Solve for \( n \).
- Take the square root of both sides:
\[
n + \frac{1}{2} = \pm \sqrt{\frac{121}{4}}
\]
\[
n + \frac{1}{2} = \pm \frac{11}{2}
\]
- Solve for \( n \):
\[
n = -\frac{1}{2} + \frac{11}{2} \quad \text{or} \quad n = -\frac{1}{2} - \frac{11}{2}
\]
\[
n = \frac{10}{2} \quad \text{or} \quad n = -\frac{12}{2}
\]
\[
n = 5 \quad \text{or} \quad n = -6
\]

#### Final Answer:
\[
\boxed{n = 5 \text{ or } n = -6}
\]

---

5) \( u(u + 6) = 3 \)

#### Step 1: Expand and rewrite the equation.
\[ u^2 + 6u = 3 \]

#### Step 2: Move the constant term to the right side.
\[ u^2 + 6u = 3 \]

#### Step 3: Complete the square on the left side.
- Take the coefficient of \( u \), which is \(6\), divide it by 2, and square it:
\[
\left(\frac{6}{2}\right)^2 = 3^2 = 9
\]
- Add this square to both sides:
\[
u^2 + 6u + 9 = 3 + 9
\]
\[
u^2 + 6u + 9 = 12
\]

#### Step 4: Write the left side as a perfect square.
\[ (u + 3)^2 = 12 \]

#### Step 5: Solve for \( u \).
- Take the square root of both sides:
\[
u + 3 = \pm \sqrt{12}
\]
\[
u + 3 = \pm 2\sqrt{3}
\]
- Solve for \( u \):
\[
u = -3 + 2\sqrt{3} \quad \text{or} \quad u = -3 - 2\sqrt{3}
\]

#### Final Answer:
\[
\boxed{u = -3 + 2\sqrt{3} \text{ or } u = -3 - 2\sqrt{3}}
\]

---

6) \( 35 = g^2 - 2g \)

#### Step 1: Move all terms to one side.
\[ g^2 - 2g - 35 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ g^2 - 2g = 35 \]

#### Step 3: Complete the square on the left side.
- Take the coefficient of \( g \), which is \(-2\), divide it by 2, and square it:
\[
\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1
\]
- Add this square to both sides:
\[
g^2 - 2g + 1 = 35 + 1
\]
\[
g^2 - 2g + 1 = 36
\]

#### Step 4: Write the left side as a perfect square.
\[ (g - 1)^2 = 36 \]

#### Step 5: Solve for \( g \).
- Take the square root of both sides:
\[
g - 1 = \pm \sqrt{36}
\]
\[
g - 1 = \pm 6
\]
- Solve for \( g \):
\[
g = 1 + 6 \quad \text{or} \quad g = 1 - 6
\]
\[
g = 7 \quad \text{or} \quad g = -5
\]

#### Final Answer:
\[
\boxed{g = 7 \text{ or } g = -5}
\]

---

7) \( 7y = y^2 + 12 \)

#### Step 1: Rearrange the equation.
\[ y^2 - 7y + 12 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ y^2 - 7y = -12 \]

#### Step 3: Complete the square on the left side.
- Take the coefficient of \( y \), which is \(-7\), divide it by 2, and square it:
\[
\left(\frac{-7}{2}\right)^2 = \left(-\frac{7}{2}\right)^2 = \frac{49}{4}
\]
- Add this square to both sides:
\[
y^2 - 7y + \frac{49}{4} = -12 + \frac{49}{4}
\]
\[
y^2 - 7y + \frac{49}{4} = -\frac{48}{4} + \frac{49}{4}
\]
\[
y^2 - 7y + \frac{49}{4} = \frac{1}{4}
\]

#### Step 4: Write the left side as a perfect square.
\[ \left(y - \frac{7}{2}\right)^2 = \frac{1}{4} \]

#### Step 5: Solve for \( y \).
- Take the square root of both sides:
\[
y - \frac{7}{2} = \pm \sqrt{\frac{1}{4}}
\]
\[
y - \frac{7}{2} = \pm \frac{1}{2}
\]
- Solve for \( y \):
\[
y = \frac{7}{2} + \frac{1}{2} \quad \text{or} \quad y = \frac{7}{2} - \frac{1}{2}
\]
\[
y = \frac{8}{2} \quad \text{or} \quad y = \frac{6}{2}
\]
\[
y = 4 \quad \text{or} \quad y = 3
\]

#### Final Answer:
\[
\boxed{y = 4 \text{ or } y = 3}
\]

---

8) \( t^2 - 7t + 6 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ t^2 - 7t = -6 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( t \), which is \(-7\), divide it by 2, and square it:
\[
\left(\frac{-7}{2}\right)^2 = \left(-\frac{7}{2}\right)^2 = \frac{49}{4}
\]
- Add this square to both sides:
\[
t^2 - 7t + \frac{49}{4} = -6 + \frac{49}{4}
\]
\[
t^2 - 7t + \frac{49}{4} = -\frac{24}{4} + \frac{49}{4}
\]
\[
t^2 - 7t + \frac{49}{4} = \frac{25}{4}
\]

#### Step 3: Write the left side as a perfect square.
\[ \left(t - \frac{7}{2}\right)^2 = \frac{25}{4} \]

#### Step 4: Solve for \( t \).
- Take the square root of both sides:
\[
t - \frac{7}{2} = \pm \sqrt{\frac{25}{4}}
\]
\[
t - \frac{7}{2} = \pm \frac{5}{2}
\]
- Solve for \( t \):
\[
t = \frac{7}{2} + \frac{5}{2} \quad \text{or} \quad t = \frac{7}{2} - \frac{5}{2}
\]
\[
t = \frac{12}{2} \quad \text{or} \quad t = \frac{2}{2}
\]
\[
t = 6 \quad \text{or} \quad t = 1
\]

#### Final Answer:
\[
\boxed{t = 6 \text{ or } t = 1}
\]

---

9) \( m^2 + 8m + 10 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ m^2 + 8m = -10 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( m \), which is \(8\), divide it by 2, and square it:
\[
\left(\frac{8}{2}\right)^2 = 4^2 = 16
\]
- Add this square to both sides:
\[
m^2 + 8m + 16 = -10 + 16
\]
\[
m^2 + 8m + 16 = 6
\]

#### Step 3: Write the left side as a perfect square.
\[ (m + 4)^2 = 6 \]

#### Step 4: Solve for \( m \).
- Take the square root of both sides:
\[
m + 4 = \pm \sqrt{6}
\]
- Solve for \( m \):
\[
m = -4 + \sqrt{6} \quad \text{or} \quad m = -4 - \sqrt{6}
\]

#### Final Answer:
\[
\boxed{m = -4 + \sqrt{6} \text{ or } m = -4 - \sqrt{6}}
\]

---

10) \( h^2 - 4h - 32 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ h^2 - 4h = 32 \]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( h \), which is \(-4\), divide it by 2, and square it:
\[
\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4
\]
- Add this square to both sides:
\[
h^2 - 4h + 4 = 32 + 4
\]
\[
h^2 - 4h + 4 = 36
\]

#### Step 3: Write the left side as a perfect square.
\[ (h - 2)^2 = 36 \]

#### Step 4: Solve for \( h \).
- Take the square root of both sides:
\[
h - 2 = \pm \sqrt{36}
\]
\[
h - 2 = \pm 6
\]
- Solve for \( h \):
\[
h = 2 + 6 \quad \text{or} \quad h = 2 - 6
\]
\[
h = 8 \quad \text{or} \quad h = -4
\]

#### Final Answer:
\[
\boxed{h = 8 \text{ or } h = -4}
\]

---

Final Answers for All Equations:

1. \( \boxed{k = 4 \text{ or } k = -2} \)
2. \( \boxed{x = 1 + 5i \text{ or } x = 1 - 5i} \)
3. \( \boxed{p = 2 \text{ or } p = -1} \)
4. \( \boxed{n = 5 \text{ or } n = -6} \)
5. \( \boxed{u = -3 + 2\sqrt{3} \text{ or } u = -3 - 2\sqrt{3}} \)
6. \( \boxed{g = 7 \text{ or } g = -5} \)
7. \( \boxed{y = 4 \text{ or } y = 3} \)
8. \( \boxed{t = 6 \text{ or } t = 1} \)
9. \( \boxed{m = -4 + \sqrt{6} \text{ or } m = -4 - \sqrt{6}} \)
10. \( \boxed{h = 8 \text{ or } h = -4} \)