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Problem: Solving Quadratic Equations by Factorising

The task involves solving quadratic equations using factorisation. Let's break it down into Section A and Section B.

---

Section A: Find the solutions to the following equations

These equations are already factored, so we can use the Zero Product Property, which states that if the product of two factors is zero, then at least one of the factors must be zero.

#### 1) \((x + 2)(x + 3) = 0\)
- Set each factor equal to zero:
\[
x + 2 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = -2 \quad \text{or} \quad x = -3
\]
- Solutions: \(\boxed{-2, -3}\)

#### 2) \((4x + 3)(x - 3) = 0\)
- Set each factor equal to zero:
\[
4x + 3 = 0 \quad \text{or} \quad x - 3 = 0
\]
\[
4x = -3 \quad \Rightarrow \quad x = -\frac{3}{4}
\]
\[
x = 3
\]
- Solutions: \(\boxed{-\frac{3}{4}, 3}\)

#### 3) \((2x - 5)(3x + 2) = 0\)
- Set each factor equal to zero:
\[
2x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0
\]
\[
2x = 5 \quad \Rightarrow \quad x = \frac{5}{2}
\]
\[
3x = -2 \quad \Rightarrow \quad x = -\frac{2}{3}
\]
- Solutions: \(\boxed{\frac{5}{2}, -\frac{2}{3}}\)

#### 4) \((4x + 8)(2x - 8) = 0\)
- Set each factor equal to zero:
\[
4x + 8 = 0 \quad \text{or} \quad 2x - 8 = 0
\]
\[
4x = -8 \quad \Rightarrow \quad x = -2
\]
\[
2x = 8 \quad \Rightarrow \quad x = 4
\]
- Solutions: \(\boxed{-2, 4}\)

#### 5) \((5x - 15)(4x - 2) = 0\)
- Set each factor equal to zero:
\[
5x - 15 = 0 \quad \text{or} \quad 4x - 2 = 0
\]
\[
5x = 15 \quad \Rightarrow \quad x = 3
\]
\[
4x = 2 \quad \Rightarrow \quad x = \frac{1}{2}
\]
- Solutions: \(\boxed{3, \frac{1}{2}}\)

#### 6) \((9x - 12)(5x + 20) = 0\)
- Set each factor equal to zero:
\[
9x - 12 = 0 \quad \text{or} \quad 5x + 20 = 0
\]
\[
9x = 12 \quad \Rightarrow \quad x = \frac{12}{9} = \frac{4}{3}
\]
\[
5x = -20 \quad \Rightarrow \quad x = -4
\]
- Solutions: \(\boxed{\frac{4}{3}, -4}\)

---

Section B: Solve the following equations by factorising

These equations are not yet factored, so we need to factorise them first and then use the Zero Product Property.

#### 1) \(x^2 + 8x + 15 = 0\)
- Factorise: We need two numbers that multiply to \(15\) and add to \(8\). These numbers are \(3\) and \(5\).
\[
x^2 + 8x + 15 = (x + 3)(x + 5)
\]
- Set each factor equal to zero:
\[
x + 3 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = -3 \quad \text{or} \quad x = -5
\]
- Solutions: \(\boxed{-3, -5}\)

#### 2) \(x^2 - 7x + 12 = 0\)
- Factorise: We need two numbers that multiply to \(12\) and add to \(-7\). These numbers are \(-3\) and \(-4\).
\[
x^2 - 7x + 12 = (x - 3)(x - 4)
\]
- Set each factor equal to zero:
\[
x - 3 = 0 \quad \text{or} \quad x - 4 = 0
\]
\[
x = 3 \quad \text{or} \quad x = 4
\]
- Solutions: \(\boxed{3, 4}\)

#### 3) \(x^2 + 2x - 15 = 0\)
- Factorise: We need two numbers that multiply to \(-15\) and add to \(2\). These numbers are \(5\) and \(-3\).
\[
x^2 + 2x - 15 = (x + 5)(x - 3)
\]
- Set each factor equal to zero:
\[
x + 5 = 0 \quad \text{or} \quad x - 3 = 0
\]
\[
x = -5 \quad \text{or} \quad x = 3
\]
- Solutions: \(\boxed{-5, 3}\)

#### 4) \(x^2 - 11x + 28 = 0\)
- Factorise: We need two numbers that multiply to \(28\) and add to \(-11\). These numbers are \(-7\) and \(-4\).
\[
x^2 - 11x + 28 = (x - 7)(x - 4)
\]
- Set each factor equal to zero:
\[
x - 7 = 0 \quad \text{or} \quad x - 4 = 0
\]
\[
x = 7 \quad \text{or} \quad x = 4
\]
- Solutions: \(\boxed{7, 4}\)

#### 5) \(x^2 - x - 30 = 0\)
- Factorise: We need two numbers that multiply to \(-30\) and add to \(-1\). These numbers are \(-6\) and \(5\).
\[
x^2 - x - 30 = (x - 6)(x + 5)
\]
- Set each factor equal to zero:
\[
x - 6 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = 6 \quad \text{or} \quad x = -5
\]
- Solutions: \(\boxed{6, -5}\)

#### 6) \(x^2 + 11x - 26 = 0\)
- Factorise: We need two numbers that multiply to \(-26\) and add to \(11\). These numbers are \(13\) and \(-2\).
\[
x^2 + 11x - 26 = (x + 13)(x - 2)
\]
- Set each factor equal to zero:
\[
x + 13 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -13 \quad \text{or} \quad x = 2
\]
- Solutions: \(\boxed{-13, 2}\)

#### 7) \(x^2 - 5x - 24 = 0\)
- Factorise: We need two numbers that multiply to \(-24\) and add to \(-5\). These numbers are \(-8\) and \(3\).
\[
x^2 - 5x - 24 = (x - 8)(x + 3)
\]
- Set each factor equal to zero:
\[
x - 8 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = 8 \quad \text{or} \quad x = -3
\]
- Solutions: \(\boxed{8, -3}\)

#### 8) \(14 + x^2 + 9x = 0\)
- Rearrange: \(x^2 + 9x + 14 = 0\)
- Factorise: We need two numbers that multiply to \(14\) and add to \(9\). These numbers are \(7\) and \(2\).
\[
x^2 + 9x + 14 = (x + 7)(x + 2)
\]
- Set each factor equal to zero:
\[
x + 7 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = -7 \quad \text{or} \quad x = -2
\]
- Solutions: \(\boxed{-7, -2}\)

#### 9) \(7 + x^2 - 8x = -25\)
- Rearrange: \(x^2 - 8x + 32 = 0\)
- This quadratic does not factorise easily, so we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = -8\), and \(c = 32\):
\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(32)}}{2(1)}
\]
\[
x = \frac{8 \pm \sqrt{64 - 128}}{2}
\]
\[
x = \frac{8 \pm \sqrt{-64}}{2}
\]
Since the discriminant is negative, there are no real solutions.
- Solutions: \(\boxed{\text{No real solutions}}\)

#### 10) \(x^2 = 17x - 72\)
- Rearrange: \(x^2 - 17x + 72 = 0\)
- Factorise: We need two numbers that multiply to \(72\) and add to \(-17\). These numbers are \(-8\) and \(-9\).
\[
x^2 - 17x + 72 = (x - 8)(x - 9)
\]
- Set each factor equal to zero:
\[
x - 8 = 0 \quad \text{or} \quad x - 9 = 0
\]
\[
x = 8 \quad \text{or} \quad x = 9
\]
- Solutions: \(\boxed{8, 9}\)

---

Final Answers:

- Section A:
1. \(\boxed{-2, -3}\)
2. \(\boxed{-\frac{3}{4}, 3}\)
3. \(\boxed{\frac{5}{2}, -\frac{2}{3}}\)
4. \(\boxed{-2, 4}\)
5. \(\boxed{3, \frac{1}{2}}\)
6. \(\boxed{\frac{4}{3}, -4}\)

- Section B:
1. \(\boxed{-3, -5}\)
2. \(\boxed{3, 4}\)
3. \(\boxed{-5, 3}\)
4. \(\boxed{7, 4}\)
5. \(\boxed{6, -5}\)
6. \(\boxed{-13, 2}\)
7. \(\boxed{8, -3}\)
8. \(\boxed{-7, -2}\)
9. \(\boxed{\text{No real solutions}}\)
10. \(\boxed{8, 9}\)

Boxed Final Answer:
\[
\boxed{
\begin{array}{ll}
\text{Section A:} & 1. \{-2, -3\}, \, 2. \left\{-\frac{3}{4}, 3\right\}, \, 3. \left\{\frac{5}{2}, -\frac{2}{3}\right\}, \, 4. \{-2, 4\}, \, 5. \left\{3, \frac{1}{2}\right\}, \, 6. \left\{\frac{4}{3}, -4\right\} \\
\text{Section B:} & 1. \{-3, -5\}, \, 2. \{3, 4\}, \, 3. \{-5, 3\}, \, 4. \{7, 4\}, \, 5. \{6, -5\}, \, 6. \{-13, 2\}, \, 7. \{8, -3\}, \, 8. \{-7, -2\}, \, 9. \{\text{No real solutions}\}, \, 10. \{8, 9\}
\end{array}
}
\]