Solve the following quadratic inequalities, writing answers in surd form where appropriate.

To solve the given quadratic inequalities, we will follow a systematic approach for each problem. The general steps are:

1. Rewrite the inequality in standard form: \( ax^2 + bx + c \leq 0 \) or \( ax^2 + bx + c \geq 0 \).
2. Solve the corresponding quadratic equation \( ax^2 + bx + c = 0 \) to find the critical points.
3. Determine the intervals defined by the critical points.
4. Test a point in each interval to determine where the inequality holds.
5. Write the solution in interval notation or surd form as appropriate.

Let's solve each inequality step by step.

---

1. \( x^2 < 6x + 27 \)

#### Step 1: Rewrite in standard form
\[ x^2 - 6x - 27 < 0 \]

#### Step 2: Solve the quadratic equation
\[ x^2 - 6x - 27 = 0 \]
Factorize:
\[ (x - 9)(x + 3) = 0 \]
So, the roots are:
\[ x = 9 \quad \text{and} \quad x = -3 \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, -3), \quad (-3, 9), \quad (9, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, -3) \), choose \( x = -4 \):
\[ (-4)^2 - 6(-4) - 27 = 16 + 24 - 27 = 13 > 0 \]
- For \( x \in (-3, 9) \), choose \( x = 0 \):
\[ 0^2 - 6(0) - 27 = -27 < 0 \]
- For \( x \in (9, \infty) \), choose \( x = 10 \):
\[ 10^2 - 6(10) - 27 = 100 - 60 - 27 = 13 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 - 6x - 27 < 0 \) holds in the interval \( (-3, 9) \).

Solution:
\[ \boxed{(-3, 9)} \]

---

2. \( 2 < 3x - x^2 \)

#### Step 1: Rewrite in standard form
\[ x^2 - 3x + 2 < 0 \]

#### Step 2: Solve the quadratic equation
\[ x^2 - 3x + 2 = 0 \]
Factorize:
\[ (x - 1)(x - 2) = 0 \]
So, the roots are:
\[ x = 1 \quad \text{and} \quad x = 2 \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, 1), \quad (1, 2), \quad (2, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, 1) \), choose \( x = 0 \):
\[ 0^2 - 3(0) + 2 = 2 > 0 \]
- For \( x \in (1, 2) \), choose \( x = 1.5 \):
\[ (1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 < 0 \]
- For \( x \in (2, \infty) \), choose \( x = 3 \):
\[ 3^2 - 3(3) + 2 = 9 - 9 + 2 = 2 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 - 3x + 2 < 0 \) holds in the interval \( (1, 2) \).

Solution:
\[ \boxed{(1, 2)} \]

---

3. \( x^2 - 6 > 6 - x \)

#### Step 1: Rewrite in standard form
\[ x^2 + x - 12 > 0 \]

#### Step 2: Solve the quadratic equation
\[ x^2 + x - 12 = 0 \]
Factorize:
\[ (x + 4)(x - 3) = 0 \]
So, the roots are:
\[ x = -4 \quad \text{and} \quad x = 3 \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, -4), \quad (-4, 3), \quad (3, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, -4) \), choose \( x = -5 \):
\[ (-5)^2 + (-5) - 12 = 25 - 5 - 12 = 8 > 0 \]
- For \( x \in (-4, 3) \), choose \( x = 0 \):
\[ 0^2 + 0 - 12 = -12 < 0 \]
- For \( x \in (3, \infty) \), choose \( x = 4 \):
\[ 4^2 + 4 - 12 = 16 + 4 - 12 = 8 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 + x - 12 > 0 \) holds in the intervals \( (-\infty, -4) \) and \( (3, \infty) \).

Solution:
\[ \boxed{(-\infty, -4) \cup (3, \infty)} \]

---

4. \( x^2 \geq 5x \)

#### Step 1: Rewrite in standard form
\[ x^2 - 5x \geq 0 \]

#### Step 2: Solve the quadratic equation
\[ x^2 - 5x = 0 \]
Factorize:
\[ x(x - 5) = 0 \]
So, the roots are:
\[ x = 0 \quad \text{and} \quad x = 5 \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, 0), \quad (0, 5), \quad (5, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, 0) \), choose \( x = -1 \):
\[ (-1)^2 - 5(-1) = 1 + 5 = 6 > 0 \]
- For \( x \in (0, 5) \), choose \( x = 1 \):
\[ 1^2 - 5(1) = 1 - 5 = -4 < 0 \]
- For \( x \in (5, \infty) \), choose \( x = 6 \):
\[ 6^2 - 5(6) = 36 - 30 = 6 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 - 5x \geq 0 \) holds in the intervals \( (-\infty, 0] \) and \( [5, \infty) \).

Solution:
\[ \boxed{(-\infty, 0] \cup [5, \infty)} \]

---

5. \( x^2 + 6x - 10 \geq 0 \)

#### Step 1: Solve the quadratic equation
\[ x^2 + 6x - 10 = 0 \]
Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 6 \), \( c = -10 \):
\[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-10)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 40}}{2} = \frac{-6 \pm \sqrt{76}}{2} = \frac{-6 \pm 2\sqrt{19}}{2} = -3 \pm \sqrt{19} \]
So, the roots are:
\[ x = -3 + \sqrt{19} \quad \text{and} \quad x = -3 - \sqrt{19} \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, -3 - \sqrt{19}), \quad (-3 - \sqrt{19}, -3 + \sqrt{19}), \quad (-3 + \sqrt{19}, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, -3 - \sqrt{19}) \), choose \( x = -10 \):
\[ (-10)^2 + 6(-10) - 10 = 100 - 60 - 10 = 30 > 0 \]
- For \( x \in (-3 - \sqrt{19}, -3 + \sqrt{19}) \), choose \( x = 0 \):
\[ 0^2 + 6(0) - 10 = -10 < 0 \]
- For \( x \in (-3 + \sqrt{19}, \infty) \), choose \( x = 10 \):
\[ 10^2 + 6(10) - 10 = 100 + 60 - 10 = 150 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 + 6x - 10 \geq 0 \) holds in the intervals \( (-\infty, -3 - \sqrt{19}] \) and \( [-3 + \sqrt{19}, \infty) \).

Solution:
\[ \boxed{(-\infty, -3 - \sqrt{19}] \cup [-3 + \sqrt{19}, \infty)} \]

---

6. \( x^2 - 2x - 9 > 0 \)

#### Step 1: Solve the quadratic equation
\[ x^2 - 2x - 9 = 0 \]
Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), \( c = -9 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-9)}}{2(1)} = \frac{2 \pm \sqrt{4 + 36}}{2} = \frac{2 \pm \sqrt{40}}{2} = \frac{2 \pm 2\sqrt{10}}{2} = 1 \pm \sqrt{10} \]
So, the roots are:
\[ x = 1 + \sqrt{10} \quad \text{and} \quad x = 1 - \sqrt{10} \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ (-\infty, 1 - \sqrt{10}), \quad (1 - \sqrt{10}, 1 + \sqrt{10}), \quad (1 + \sqrt{10}, \infty) \]

#### Step 4: Test points in each interval
- For \( x \in (-\infty, 1 - \sqrt{10}) \), choose \( x = -10 \):
\[ (-10)^2 - 2(-10) - 9 = 100 + 20 - 9 = 111 > 0 \]
- For \( x \in (1 - \sqrt{10}, 1 + \sqrt{10}) \), choose \( x = 0 \):
\[ 0^2 - 2(0) - 9 = -9 < 0 \]
- For \( x \in (1 + \sqrt{10}, \infty) \), choose \( x = 10 \):
\[ 10^2 - 2(10) - 9 = 100 - 20 - 9 = 71 > 0 \]

#### Step 5: Write the solution
The inequality \( x^2 - 2x - 9 > 0 \) holds in the intervals \( (-\infty, 1 - \sqrt{10}) \) and \( (1 + \sqrt{10}, \infty) \).

Solution:
\[ \boxed{(-\infty, 1 - \sqrt{10}) \cup (1 + \sqrt{10}, \infty)} \]

---

7. \( 2x^2 \leq 7 - 2x \)

#### Step 1: Rewrite in standard form
\[ 2x^2 + 2x - 7 \leq 0 \]

#### Step 2: Solve the quadratic equation
\[ 2x^2 + 2x - 7 = 0 \]
Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = 2 \), \( c = -7 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-7)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 56}}{4} = \frac{-2 \pm \sqrt{60}}{4} = \frac{-2 \pm 2\sqrt{15}}{4} = \frac{-1 \pm \sqrt{15}}{2} \]
So, the roots are:
\[ x = \frac{-1 + \sqrt{15}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{15}}{2} \]

#### Step 3: Determine intervals
The critical points divide the real line into three intervals:
\[ \left( -\infty, \frac{-1 - \sqrt{15}}{2} \right), \quad \left( \frac{-1 - \sqrt{15}}{2}, \frac{-1 + \sqrt{15}}{2} \right), \quad \left( \frac{-1 + \sqrt{15}}{2}, \infty \right) \]

#### Step 4: Test points in each interval
- For \( x \in \left( -\infty, \frac{-1 - \sqrt{15}}{2} \right) \), choose \( x = -3 \):
\[ 2(-3)^2 + 2(-3) - 7 = 18 - 6 - 7 = 5 > 0 \]
- For \( x \in \left( \frac{-1 - \sqrt{15}}{2}, \frac{-1 + \sqrt{15}}{2} \right) \), choose \( x = 0 \):
\[ 2(0)^2 + 2(0) - 7 = -7 < 0 \]
- For \( x \in \left( \frac{-1 + \sqrt{15}}{2}, \infty \right) \), choose \( x = 1 \):
\[ 2(1)^2 + 2(1) - 7 = 2 + 2 - 7 = -3 < 0 \]

#### Step 5: Write the solution
The inequality \( 2x^2 + 2x - 7 \leq 0 \) holds in the interval \( \left[ \frac{-1 - \sqrt{15}}{2}, \frac{-1 + \sqrt{15}}{2} \right] \).

Solution:
\[ \boxed{\left[ \frac{-1 - \sqrt{15}}{2}, \frac{-1 + \sqrt{15}}{2} \right]} \]

---

8. \( -1 < -3x - 5x^2 \)

#### Step 1: Rewrite in standard form
\[ 5x^2 + 3x + 1 > 0 \]

#### Step 2: Solve the quadratic equation
\[ 5x^2 + 3x + 1 = 0 \]
Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = 3 \), \( c = 1 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4(5)(1)}}{2(5)} = \frac{-3 \pm \sqrt{9 - 20}}{10} = \frac{-3 \pm \sqrt{-11}}{10} \]
Since the discriminant is negative (\( \sqrt{-11} \)), there are no real roots. The quadratic \( 5x^2 + 3x + 1 \) does not cross the x-axis.

#### Step 3: Determine the sign of the quadratic
The quadratic \( 5x^2 + 3x + 1 \) opens upwards (since the coefficient of \( x^2 \) is positive). Since it does not cross the x-axis, it is always positive.

#### Step 4: Write the solution
The inequality \( 5x^2 + 3x + 1 > 0 \) holds for all real \( x \).

Solution:
\[ \boxed{(-\infty, \infty)} \]

---

Final Answers:

1. \( \boxed{(-3, 9)} \)
2. \( \boxed{(1, 2)} \)
3. \( \boxed{(-\infty, -4) \cup (3, \infty)} \)
4. \( \boxed{(-\infty, 0] \cup [5, \infty)} \)
5. \( \boxed{(-\infty, -3 - \sqrt{19}] \cup [-3 + \sqrt{19}, \infty)} \)
6. \( \boxed{(-\infty, 1 - \sqrt{10}) \cup (1 + \sqrt{10}, \infty)} \)
7. \( \boxed{\left[ \frac{-1 - \sqrt{15}}{2}, \frac{-1 + \sqrt{15}}{2} \right]} \)
8. \( \boxed{(-\infty, \infty)} \)