Sure, let's solve each equation step by step.

### a) \((2x-3)^2 - 4x^2 = -15\)

1. Expand \((2x-3)^2\):
\[(2x-3)^2 = 4x^2 - 12x + 9\]

2. Substitute back into the equation:
\[4x^2 - 12x + 9 - 4x^2 = -15\]

3. Simplify:
\[-12x + 9 = -15\]

4. Solve for \(x\):
\[-12x = -15 - 9\]
\[-12x = -24\]
\[x = 2\]

### b) \((x-2)^2 - x^2 + 16 = 0\)

1. Expand \((x-2)^2\):
\[(x-2)^2 = x^2 - 4x + 4\]

2. Substitute back into the equation:
\[x^2 - 4x + 4 - x^2 + 16 = 0\]

3. Simplify:
\[-4x + 20 = 0\]

4. Solve for \(x\):
\[-4x = -20\]
\[x = 5\]

### c) \(9x^2 - 16 - (4 - 3x)(x - 4) = 0\)

1. Expand \((4 - 3x)(x - 4)\):
\[(4 - 3x)(x - 4) = 4x - 16 - 3x^2 + 12x = -3x^2 + 16x - 16\]

2. Substitute back into the equation:
\[9x^2 - 16 + 3x^2 - 16x + 16 = 0\]

3. Simplify:
\[12x^2 - 16x = 0\]

4. Factor:
\[4x(3x - 4) = 0\]

5. Set each factor to zero:
\[4x = 0 \quad \text{or} \quad 3x - 4 = 0\]

Solutions:
\[x = 0\]
\[x = \frac{4}{3}\]

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

1. Simplify:
\[3x + 12 - x^2 - 4x = 0\]

2. Rearrange terms:
\[-x^2 - x + 12 = 0\]

3. Multiply all terms by -1:
\[x^2 + x - 12 = 0\]

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

5. Solve for \(x\):
\[x = -4 \quad \text{or} \quad x = 3\]

### e) \(\left(2x + \frac{1}{2}\right)^2 - (1 - 2x)^2 = 2\)

1. Expand \(\left(2x + \frac{1}{2}\right)^2\) and \((1 - 2x)^2\):
\[\left(2x + \frac{1}{2}\right)^2 = 4x^2 + 2x + \frac{1}{4}\]
\[(1 - 2x)^2 = 1 - 4x + 4x^2\]

2. Substitute back into the equation:
\[4x^2 + 2x + \frac{1}{4} - 1 + 4x - 4x^2 = 2\]

3. Simplify:
\[6x - \frac{3}{4} = 2\]

4. Solve for \(x\):
\[6x = 2 + \frac{3}{4}\]
\[6x = \frac{8}{4} + \frac{3}{4}\]
\[6x = \frac{11}{4}\]
\[x = \frac{11}{24}\]

### f) \(x^3 - 3x^2 + 3x - 28 = 0\)

This equation is a cubic and is more complex to factor directly. Generally, try using the Rational Root Theorem and synthetic division.

1. Test possible rational roots: \(\pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28\).

2. Through testing and/or using software, find that \(x = 4\) is a root.

3. Perform synthetic division to simplify:
\((x-4)(x^2 + x + 7) = 0\)

4. Solve:
\[x = 4\]

The quadratic \(x^2 + x + 7 = 0\) has no real roots (discriminant is negative).

So the real solution is:
\[x = 4\]

If you need solutions for specific equations or further clarification, let me know!