Let's solve each of the given equations step by step.

### 24) \((x^2-4) + (x-2)(3-2x) = 0\)

First, simplify the equation:
\[ x^2 - 4 + (x - 2)(3 - 2x) = 0 \]

Expand the second term:
\[ (x - 2)(3 - 2x) = 3x - 2x^2 - 6 + 4x = -2x^2 + 7x - 6 \]

Now substitute back:
\[ x^2 - 4 - 2x^2 + 7x - 6 = 0 \]
\[ -x^2 + 7x - 10 = 0 \]

Multiply by -1 to simplify:
\[ x^2 - 7x + 10 = 0 \]

Factorize:
\[ (x - 2)(x - 5) = 0 \]

So, the solutions are:
\[ x = 2 \quad \text{or} \quad x = 5 \]

### 25) \((3x-2)(x+1) = x^2 - 1\)

First, expand the left-hand side:
\[ (3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2 \]

Now set the equation:
\[ 3x^2 + x - 2 = x^2 - 1 \]

Move all terms to one side:
\[ 3x^2 + x - 2 - x^2 + 1 = 0 \]
\[ 2x^2 + x - 1 = 0 \]

Factorize:
\[ (2x - 1)(x + 1) = 0 \]

So, the solutions are:
\[ x = \frac{1}{2} \quad \text{or} \quad x = -1 \]

### 26) \(2(4-x) + 3x(x-4) = 0\)

First, expand the equation:
\[ 2(4 - x) + 3x(x - 4) = 0 \]
\[ 8 - 2x + 3x^2 - 12x = 0 \]
\[ 3x^2 - 14x + 8 = 0 \]

Factorize:
\[ (3x - 2)(x - 4) = 0 \]

So, the solutions are:
\[ x = \frac{2}{3} \quad \text{or} \quad x = 4 \]

### 27) \((2x+3)^2 = (x-7)^2\)

Take the square root of both sides:
\[ 2x + 3 = x - 7 \quad \text{or} \quad 2x + 3 = -(x - 7) \]

Solve each equation:
1. \(2x + 3 = x - 7\)
\[ 2x - x = -7 - 3 \]
\[ x = -10 \]

2. \(2x + 3 = -x + 7\)
\[ 2x + x = 7 - 3 \]
\[ 3x = 4 \]
\[ x = \frac{4}{3} \]

So, the solutions are:
\[ x = -10 \quad \text{or} \quad x = \frac{4}{3} \]

### 28) \(4x^2 - 1 = (2x+1)(3x-5)\)

First, expand the right-hand side:
\[ (2x + 1)(3x - 5) = 6x^2 - 10x + 3x - 5 = 6x^2 - 7x - 5 \]

Now set the equation:
\[ 4x^2 - 1 = 6x^2 - 7x - 5 \]

Move all terms to one side:
\[ 4x^2 - 1 - 6x^2 + 7x + 5 = 0 \]
\[ -2x^2 + 7x + 4 = 0 \]

Multiply by -1 to simplify:
\[ 2x^2 - 7x - 4 = 0 \]

Factorize:
\[ (2x + 1)(x - 4) = 0 \]

So, the solutions are:
\[ x = -\frac{1}{2} \quad \text{or} \quad x = 4 \]

### 29) \(4x^2 - 9 = (x+7)(2x+3)\)

First, expand the right-hand side:
\[ (x + 7)(2x + 3) = 2x^2 + 3x + 14x + 21 = 2x^2 + 17x + 21 \]

Now set the equation:
\[ 4x^2 - 9 = 2x^2 + 17x + 21 \]

Move all terms to one side:
\[ 4x^2 - 9 - 2x^2 - 17x - 21 = 0 \]
\[ 2x^2 - 17x - 30 = 0 \]

Factorize:
\[ (2x + 3)(x - 10) = 0 \]

So, the solutions are:
\[ x = -\frac{3}{2} \quad \text{or} \quad x = 10 \]

### 30) \(9x^2 - 4 - (3x+2)(x-1) = 0\)

First, expand the term:
\[ (3x + 2)(x - 1) = 3x^2 - 3x + 2x - 2 = 3x^2 - x - 2 \]

Now set the equation:
\[ 9x^2 - 4 - 3x^2 + x + 2 = 0 \]
\[ 6x^2 + x - 2 = 0 \]

Factorize:
\[ (3x - 1)(2x + 2) = 0 \]

So, the solutions are:
\[ x = \frac{1}{3} \quad \text{or} \quad x = -1 \]