Solving Quadratics by Factoring Mixed Level worksheet with an option bank and instructions to match solutions.

To solve the quadratic equations and match them to the correct options in the bank, we will follow these steps for each equation:

Step-by-Step Solution:

#### 1. Solve \( x^2 + x - 20 = 10 \)

1. Rewrite the equation in standard form:
\[
x^2 + x - 20 - 10 = 0 \implies x^2 + x - 30 = 0
\]

2. Factor the quadratic equation:
We need two numbers that multiply to \(-30\) and add to \(1\). These numbers are \(6\) and \(-5\).
\[
x^2 + x - 30 = (x + 6)(x - 5) = 0
\]

3. Solve for \(x\):
\[
x + 6 = 0 \quad \text{or} \quad x - 5 = 0
\]
\[
x = -6 \quad \text{or} \quad x = 5
\]

4. Match the solution to the option bank:
The solution is \(\{-6, 5\}\), which corresponds to option F.

---

#### 2. Solve \( 6x^2 + 11x + 4 = 0 \)

1. Factor the quadratic equation:
We need two numbers that multiply to \(6 \times 4 = 24\) and add to \(11\). These numbers are \(8\) and \(3\).
Rewrite the middle term using these numbers:
\[
6x^2 + 8x + 3x + 4 = 0
\]
Factor by grouping:
\[
2x(3x + 4) + 1(3x + 4) = 0 \implies (2x + 1)(3x + 4) = 0
\]

2. Solve for \(x\):
\[
2x + 1 = 0 \quad \text{or} \quad 3x + 4 = 0
\]
\[
x = -\frac{1}{2} \quad \text{or} \quad x = -\frac{4}{3}
\]

3. Match the solution to the option bank:
The solution is \(\left\{-\frac{4}{3}, -\frac{1}{2}\right\}\), which corresponds to option C.

---

#### 3. Solve \( x^2 - 49 = 0 \)

1. Rewrite the equation:
This is a difference of squares:
\[
x^2 - 49 = (x + 7)(x - 7) = 0
\]

2. Solve for \(x\):
\[
x + 7 = 0 \quad \text{or} \quad x - 7 = 0
\]
\[
x = -7 \quad \text{or} \quad x = 7
\]

3. Match the solution to the option bank:
The solution is \(\{-7, 7\}\), which corresponds to option J.

---

#### 4. Solve \( 4 = -2x^2 + 3x + 9 \)

1. Rewrite the equation in standard form:
\[
-2x^2 + 3x + 9 - 4 = 0 \implies -2x^2 + 3x + 5 = 0
\]
Multiply through by \(-1\) to simplify:
\[
2x^2 - 3x - 5 = 0
\]

2. Factor the quadratic equation:
We need two numbers that multiply to \(2 \times (-5) = -10\) and add to \(-3\). These numbers are \(-5\) and \(2\).
Rewrite the middle term using these numbers:
\[
2x^2 - 5x + 2x - 5 = 0
\]
Factor by grouping:
\[
x(2x - 5) + 1(2x - 5) = 0 \implies (x + 1)(2x - 5) = 0
\]

3. Solve for \(x\):
\[
x + 1 = 0 \quad \text{or} \quad 2x - 5 = 0
\]
\[
x = -1 \quad \text{or} \quad x = \frac{5}{2}
\]

4. Match the solution to the option bank:
The solution is \(\left\{-1, \frac{5}{2}\right\}\), which corresponds to option D.

---

#### 5. Solve \( x^2 + 2x - 9 = 2x \)

1. Rewrite the equation in standard form:
\[
x^2 + 2x - 9 - 2x = 0 \implies x^2 - 9 = 0
\]

2. Factor the quadratic equation:
This is a difference of squares:
\[
x^2 - 9 = (x + 3)(x - 3) = 0
\]

3. Solve for \(x\):
\[
x + 3 = 0 \quad \text{or} \quad x - 3 = 0
\]
\[
x = -3 \quad \text{or} \quad x = 3
\]

4. Match the solution to the option bank:
The solution is \(\{-3, 3\}\), which corresponds to option I.

---

#### 6. Solve \( 5x^2 - 30x + 5 = -4x \)

1. Rewrite the equation in standard form:
\[
5x^2 - 30x + 5 + 4x = 0 \implies 5x^2 - 26x + 5 = 0
\]

2. Factor the quadratic equation:
We need two numbers that multiply to \(5 \times 5 = 25\) and add to \(-26\). These numbers are \(-1\) and \(-25\).
Rewrite the middle term using these numbers:
\[
5x^2 - 25x - x + 5 = 0
\]
Factor by grouping:
\[
5x(x - 5) - 1(x - 5) = 0 \implies (5x - 1)(x - 5) = 0
\]

3. Solve for \(x\):
\[
5x - 1 = 0 \quad \text{or} \quad x - 5 = 0
\]
\[
x = \frac{1}{5} \quad \text{or} \quad x = 5
\]

4. Match the solution to the option bank:
The solution is \(\left\{\frac{1}{5}, 5\right\}\), which corresponds to option B.

---

Final Answers:

1. F
2. C
3. J
4. D
5. I
6. B

Boxed Answer:

\[
\boxed{F, C, J, D, I, B}
\]