Students solve quadratic equations using the completing the square method and color the answers according to the provided color key.

To solve the quadratic equations by completing the square, we will follow these steps for each equation:

1. Rewrite the equation in standard form: Ensure the equation is in the form \( ax^2 + bx + c = 0 \).
2. Move the constant term to the other side: Isolate the \( x \)-terms on one side.
3. Complete the square: Add and subtract the square of half the coefficient of \( x \) inside the equation.
4. Solve for \( x \): Use the square root property to find the roots.

Let's solve each equation step by step.

---

Equation 1: \( x^2 + 8x - 14 = 6 \)

1. Rewrite the equation:
\[
x^2 + 8x - 14 - 6 = 0 \implies x^2 + 8x - 20 = 0
\]

2. Move the constant term:
\[
x^2 + 8x = 20
\]

3. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{8}{2} = 4 \).
- Square it: \( 4^2 = 16 \).
- Add and subtract 16:
\[
x^2 + 8x + 16 - 16 = 20 \implies (x + 4)^2 - 16 = 20
\]
- Simplify:
\[
(x + 4)^2 = 36
\]

4. Solve for \( x \):
\[
x + 4 = \pm \sqrt{36} \implies x + 4 = \pm 6
\]
- \( x + 4 = 6 \implies x = 2 \)
- \( x + 4 = -6 \implies x = -10 \)

Roots: \( x = 2 \) and \( x = -10 \)

---

Equation 2: \( x^2 + 5x = -4 \)

1. Move the constant term:
\[
x^2 + 5x + 4 = 0
\]

2. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{5}{2} \).
- Square it: \( \left( \frac{5}{2} \right)^2 = \frac{25}{4} \).
- Add and subtract \( \frac{25}{4} \):
\[
x^2 + 5x + \frac{25}{4} - \frac{25}{4} = -4 \implies \left( x + \frac{5}{2} \right)^2 - \frac{25}{4} = -4
\]
- Simplify:
\[
\left( x + \frac{5}{2} \right)^2 = \frac{25}{4} - \frac{16}{4} = \frac{9}{4}
\]

3. Solve for \( x \):
\[
x + \frac{5}{2} = \pm \sqrt{\frac{9}{4}} \implies x + \frac{5}{2} = \pm \frac{3}{2}
\]
- \( x + \frac{5}{2} = \frac{3}{2} \implies x = \frac{3}{2} - \frac{5}{2} = -1 \)
- \( x + \frac{5}{2} = -\frac{3}{2} \implies x = -\frac{3}{2} - \frac{5}{2} = -4 \)

Roots: \( x = -1 \) and \( x = -4 \)

---

Equation 3: \( x^2 + 7x + 10 = 0 \)

1. Move the constant term:
\[
x^2 + 7x = -10
\]

2. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{7}{2} \).
- Square it: \( \left( \frac{7}{2} \right)^2 = \frac{49}{4} \).
- Add and subtract \( \frac{49}{4} \):
\[
x^2 + 7x + \frac{49}{4} - \frac{49}{4} = -10 \implies \left( x + \frac{7}{2} \right)^2 - \frac{49}{4} = -10
\]
- Simplify:
\[
\left( x + \frac{7}{2} \right)^2 = \frac{49}{4} - \frac{40}{4} = \frac{9}{4}
\]

3. Solve for \( x \):
\[
x + \frac{7}{2} = \pm \sqrt{\frac{9}{4}} \implies x + \frac{7}{2} = \pm \frac{3}{2}
\]
- \( x + \frac{7}{2} = \frac{3}{2} \implies x = \frac{3}{2} - \frac{7}{2} = -2 \)
- \( x + \frac{7}{2} = -\frac{3}{2} \implies x = -\frac{3}{2} - \frac{7}{2} = -5 \)

Roots: \( x = -2 \) and \( x = -5 \)

---

Equation 4: \( x^2 + 12x + 16 = -19 \)

1. Move the constant term:
\[
x^2 + 12x + 16 + 19 = 0 \implies x^2 + 12x + 35 = 0
\]

2. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{12}{2} = 6 \).
- Square it: \( 6^2 = 36 \).
- Add and subtract 36:
\[
x^2 + 12x + 36 - 36 + 35 = 0 \implies (x + 6)^2 - 1 = 0
\]
- Simplify:
\[
(x + 6)^2 = 1
\]

3. Solve for \( x \):
\[
x + 6 = \pm \sqrt{1} \implies x + 6 = \pm 1
\]
- \( x + 6 = 1 \implies x = -5 \)
- \( x + 6 = -1 \implies x = -7 \)

Roots: \( x = -5 \) and \( x = -7 \)

---

Equation 5: \( x^2 - 4x - 91 = 7 \)

1. Rewrite the equation:
\[
x^2 - 4x - 91 - 7 = 0 \implies x^2 - 4x - 98 = 0
\]

2. Move the constant term:
\[
x^2 - 4x = 98
\]

3. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{-4}{2} = -2 \).
- Square it: \( (-2)^2 = 4 \).
- Add and subtract 4:
\[
x^2 - 4x + 4 - 4 = 98 \implies (x - 2)^2 - 4 = 98
\]
- Simplify:
\[
(x - 2)^2 = 102
\]

4. Solve for \( x \):
\[
x - 2 = \pm \sqrt{102}
\]
- \( x - 2 = \sqrt{102} \implies x = 2 + \sqrt{102} \)
- \( x - 2 = -\sqrt{102} \implies x = 2 - \sqrt{102} \)

Roots: \( x = 2 + \sqrt{102} \) and \( x = 2 - \sqrt{102} \)

---

Equation 6: \( x^2 - 4x + 1 = -5 \)

1. Rewrite the equation:
\[
x^2 - 4x + 1 + 5 = 0 \implies x^2 - 4x + 6 = 0
\]

2. Move the constant term:
\[
x^2 - 4x = -6
\]

3. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{-4}{2} = -2 \).
- Square it: \( (-2)^2 = 4 \).
- Add and subtract 4:
\[
x^2 - 4x + 4 - 4 = -6 \implies (x - 2)^2 - 4 = -6
\]
- Simplify:
\[
(x - 2)^2 = -2
\]

4. Solve for \( x \):
\[
x - 2 = \pm \sqrt{-2} \implies x - 2 = \pm i\sqrt{2}
\]
- \( x - 2 = i\sqrt{2} \implies x = 2 + i\sqrt{2} \)
- \( x - 2 = -i\sqrt{2} \implies x = 2 - i\sqrt{2} \)

Roots: \( x = 2 + i\sqrt{2} \) and \( x = 2 - i\sqrt{2} \)

---

Equation 7: \( x^2 + 12x = -32 \)

1. Move the constant term:
\[
x^2 + 12x + 32 = 0
\]

2. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{12}{2} = 6 \).
- Square it: \( 6^2 = 36 \).
- Add and subtract 36:
\[
x^2 + 12x + 36 - 36 + 32 = 0 \implies (x + 6)^2 - 4 = 0
\]
- Simplify:
\[
(x + 6)^2 = 4
\]

3. Solve for \( x \):
\[
x + 6 = \pm \sqrt{4} \implies x + 6 = \pm 2
\]
- \( x + 6 = 2 \implies x = -4 \)
- \( x + 6 = -2 \implies x = -8 \)

Roots: \( x = -4 \) and \( x = -8 \)

---

Equation 8: \( 10x^2 - 45x - 12 = 43 \)

1. Rewrite the equation:
\[
10x^2 - 45x - 12 - 43 = 0 \implies 10x^2 - 45x - 55 = 0
\]

2. Divide by 5 to simplify:
\[
2x^2 - 9x - 11 = 0
\]

3. Move the constant term:
\[
2x^2 - 9x = 11
\]

4. Complete the square:
- Factor out 2 from the \( x \)-terms:
\[
2\left( x^2 - \frac{9}{2}x \right) = 11
\]
- Half of the coefficient of \( x \) is \( \frac{-9/2}{2} = -\frac{9}{4} \).
- Square it: \( \left( -\frac{9}{4} \right)^2 = \frac{81}{16} \).
- Add and subtract \( \frac{81}{16} \) inside the parentheses:
\[
2\left( x^2 - \frac{9}{2}x + \frac{81}{16} - \frac{81}{16} \right) = 11 \implies 2\left( \left( x - \frac{9}{4} \right)^2 - \frac{81}{16} \right) = 11
\]
- Distribute the 2:
\[
2\left( x - \frac{9}{4} \right)^2 - \frac{162}{16} = 11 \implies 2\left( x - \frac{9}{4} \right)^2 - \frac{81}{8} = 11
\]
- Simplify:
\[
2\left( x - \frac{9}{4} \right)^2 = 11 + \frac{81}{8} = \frac{88}{8} + \frac{81}{8} = \frac{169}{8}
\]
- Divide by 2:
\[
\left( x - \frac{9}{4} \right)^2 = \frac{169}{16}
\]

5. Solve for \( x \):
\[
x - \frac{9}{4} = \pm \sqrt{\frac{169}{16}} \implies x - \frac{9}{4} = \pm \frac{13}{4}
\]
- \( x - \frac{9}{4} = \frac{13}{4} \implies x = \frac{22}{4} = \frac{11}{2} \)
- \( x - \frac{9}{4} = -\frac{13}{4} \implies x = -\frac{4}{4} = -1 \)

Roots: \( x = \frac{11}{2} \) and \( x = -1 \)

---

Equation 9: \( x^2 = 12 - 4x \)

1. Rewrite the equation:
\[
x^2 + 4x - 12 = 0
\]

2. Move the constant term:
\[
x^2 + 4x = 12
\]

3. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{4}{2} = 2 \).
- Square it: \( 2^2 = 4 \).
- Add and subtract 4:
\[
x^2 + 4x + 4 - 4 = 12 \implies (x + 2)^2 - 4 = 12
\]
- Simplify:
\[
(x + 2)^2 = 16
\]

4. Solve for \( x \):
\[
x + 2 = \pm \sqrt{16} \implies x + 2 = \pm 4
\]
- \( x + 2 = 4 \implies x = 2 \)
- \( x + 2 = -4 \implies x = -6 \)

Roots: \( x = 2 \) and \( x = -6 \)

---

Equation 10: \( 3x^2 = -4 + 8x \)

1. Rewrite the equation:
\[
3x^2 - 8x + 4 = 0
\]

2. Divide by 3 to simplify:
\[
x^2 - \frac{8}{3}x + \frac{4}{3} = 0
\]

3. Move the constant term:
\[
x^2 - \frac{8}{3}x = -\frac{4}{3}
\]

4. Complete the square:
- Half of the coefficient of \( x \) is \( \frac{-8/3}{2} = -\frac{4}{3} \).
- Square it: \( \left( -\frac{4}{3} \right)^2 = \frac{16}{9} \).
- Add and subtract \( \frac{16}{9} \):
\[
x^2 - \frac{8}{3}x + \frac{16}{9} - \frac{16}{9} = -\frac{4}{3} \implies \left( x - \frac{4}{3} \right)^2 - \frac{16}{9} = -\frac{4}{3}
\]
- Simplify:
\[
\left( x - \frac{4}{3} \right)^2 = \frac{16}{9} - \frac{12}{9} = \frac{4}{9}
\]

5. Solve for \( x \):
\[
x - \frac{4}{3} = \pm \sqrt{\frac{4}{9}} \implies x - \frac{4}{3} = \pm \frac{2}{3}
\]
- \( x - \frac{4}{3} = \frac{2}{3} \implies x = \frac{6}{3} = 2 \)
- \( x - \frac{4}{3} = -\frac{2}{3} \implies x = \frac{2}{3} \)

Roots: \( x = 2 \) and \( x = \frac{2}{3} \)

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Final Answers

\[
\boxed{
\begin{array}{ccc}
\text{Equation} & \text{Roots} & \text{Color} \\
1. & x = 2, x = -10 & \text{Orange} \\
2. & x = -1, x = -4 & \text{Green} \\
3. & x = -2, x = -5 & \text{Purple} \\
4. & x = -5, x = -7 & \text{Blue} \\
5. & x = 2 + \sqrt{102}, x = 2 - \sqrt{102} & \text{Purple} \\
6. & x = 2 + i\sqrt{2}, x = 2 - i\sqrt{2} & \text{Teal} \\
7. & x = -4, x = -8 & \text{Pink} \\
8. & x = \frac{11}{2}, x = -1 & \text{Orange} \\
9. & x = 2, x = -6 & \text{Blue} \\
10. & x = 2, x = \frac{2}{3} & \text{Green} \\
\end{array}
}
\]