8. $$ \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 $$ 9. $$ (x-2)(x+1)=0 $$ 10. $$ x^{2}-4 x=0 $$ 11. $$ x^{2}-16=0 $$

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Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$x^{2}-16=0$$
- step1: Add or subtract both sides:
  $$x^{2}=16$$
- step2: Simplify the expression:
  $$x=\pm \sqrt{16}$$
- step3: Simplify the expression:
  $$x=\pm 4$$
- step4: Separate into possible cases:
  $$\begin{align}&x=4\&x=-4\end{align}$$
- step5: Rewrite:
  $$x_{1}=-4,x_{2}=4$$
Solve the equation $$ (x-2)(x+1)=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$\left(x-2\right)\left(x+1\right)=0$$
- step1: Separate into possible cases:
  $$\begin{align}&x-2=0\&x+1=0\end{align}$$
- step2: Solve the equation:
  $$\begin{align}&x=2\&x=-1\end{align}$$
- step3: Rewrite:
  $$x_{1}=-1,x_{2}=2$$
Solve the equation $$ x^{2}-4 x=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$x^{2}-4x=0$$
- step1: Factor the expression:
  $$x\left(x-4\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&x-4=0\&x=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=4\&x=0\end{align}$$
- step4: Rewrite:
  $$x_{1}=0,x_{2}=4$$
Solve the equation $$ \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{3}{4}\left(4x-12\right)+x=\frac{2}{3}\left(9x-5\right)+1$$
- step1: Simplify:
  $$4x-9=6x-\frac{7}{3}$$
- step2: Multiply both sides of the equation by LCM:
  $$\left(4x-9\right)\times 3=\left(6x-\frac{7}{3}\right)\times 3$$
- step3: Calculate:
  $$12x-27=\left(6x-\frac{7}{3}\right)\times 3$$
- step4: Calculate:
  $$12x-27=18x-7$$
- step5: Move the expression to the left side:
  $$12x-18x=-7+27$$
- step6: Add and subtract:
  $$-6x=-7+27$$
- step7: Add and subtract:
  $$-6x=20$$
- step8: Change the signs:
  $$6x=-20$$
- step9: Divide both sides:
  $$\frac{6x}{6}=\frac{-20}{6}$$
- step10: Divide the numbers:
  $$x=-\frac{10}{3}$$
Here are the solutions for each of the equations provided:

### 8. Solve $$ \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 $$

1. Distribute the terms:
   $$
   \frac{3}{4} \cdot 4x - \frac{3}{4} \cdot 12 + x = \frac{2}{3} \cdot 9x - \frac{2}{3} \cdot 5 + 1
   $$
   This simplifies to:
   $$
   3x - 9 + x = 6x - \frac{10}{3} + 1
   $$

2. Combine like terms:
   $$
   4x - 9 = 6x - \frac{10}{3} + 1
   $$

3. Rearranging gives:
   $$
   4x - 6x = -\frac{10}{3} + 1 + 9
   $$
   $$
   -2x = -\frac{10}{3} + \frac{3}{3} + \frac{27}{3}
   $$
   $$
   -2x = -\frac{10}{3} + \frac{30}{3} = \frac{20}{3}
   $$

4. Solving for $$ x $$:
   $$
   x = -\frac{10}{3}
   $$

### 9. Solve $$ (x-2)(x+1)=0 $$

Setting each factor to zero:
1. $$ x - 2 = 0 $$ gives $$ x_1 = 2 $$
2. $$ x + 1 = 0 $$ gives $$ x_2 = -1 $$

Thus, the solutions are:
$$
x_1 = 2, \quad x_2 = -1
$$

### 10. Solve $$ x^{2}-4 x=0 $$

Factoring out $$ x $$:
$$
x(x - 4) = 0
$$

Setting each factor to zero:
1. $$ x = 0 $$
2. $$ x - 4 = 0 $$ gives $$ x_2 = 4 $$

Thus, the solutions are:
$$
x_1 = 0, \quad x_2 = 4
$$

### 11. Solve $$ x^{2}-16=0 $$

Factoring as a difference of squares:
$$
(x - 4)(x + 4) = 0
$$

Setting each factor to zero:
1. $$ x - 4 = 0 $$ gives $$ x_1 = 4 $$
2. $$ x + 4 = 0 $$ gives $$ x_2 = -4 $$

Thus, the solutions are:
$$
x_1 = 4, \quad x_2 = -4
$$

### Summary of Solutions:
- **8:** $$ x = -\frac{10}{3} $$
- **9:** $$ x_1 = 2, \quad x_2 = -1 $$
- **10:** $$ x_1 = 0, \quad x_2 = 4 $$
- **11:** $$ x_1 = 4, \quad x_2 = -4 $$
The solutions are: - $$ x = -\frac{10}{3} $$ - $$ x_1 = 2, \quad x_2 = -1 $$ - $$ x_1 = 0, \quad x_2 = 4 $$ - $$ x_1 = 4, \quad x_2 = -4 $$

8. \( \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 \) 9. \( (x-2)(x+1)=0 \) 10. \( x^{2}-4 x=0 \) 11. \( x^{2}-16=0 \)

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