Let’s solve the three equations step by step:

1) Solve:

$-28x + 7x^2 = 0$

Factorize the equation:
$7x(-4 + x) = 0$

This gives two possible solutions:
$x = 0 \quad \text{or} \quad x = 4$

2) Solve:

$(x - 1)^2 = 64$

Take the square root of both sides:
$x - 1 = \pm 8$

Solve for $x$:
$x = 1 + 8 = 9 \quad \text{or} \quad x = 1 - 8 = -7$

So, the solutions are:
$x = 9 \quad \text{and} \quad x = -7$

3) Solve:

$(2x - 1)^2 + (4x + 3)^2 - 15 = 0$

Let us expand the terms:
$(2x - 1)^2 = 4x^2 - 4x + 1, \quad (4x + 3)^2 = 16x^2 + 24x + 9$

Substitute these into the equation:
$4x^2 - 4x + 1 + 16x^2 + 24x + 9 - 15 = 0$

Combine like terms:
$20x^2 + 20x - 5 = 0$

Simplify:
$4x^2 + 4x - 1 = 0$

Solve this quadratic equation using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 4$, $b = 4$, $c = -1$:
$x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-1)}}{2(4)}$ $x = \frac{-4 \pm \sqrt{16 + 16}}{8}$ $x = \frac{-4 \pm \sqrt{32}}{8}$ $x = \frac{-4 \pm 4\sqrt{2}}{8}$ $x = \frac{-1 \pm \sqrt{2}}{2}$

So, the solutions are:
$x = \frac{-1 + \sqrt{2}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{2}}{2}$

Final Solutions:

1. $x = 0, 4$
2. $x = 9, -7$
3. $x = \frac{-1 + \sqrt{2}}{2}, \frac{-1 - \sqrt{2}}{2}$

Would you like further explanation for any of these steps?

Related Questions:

1. How do you factorize more complex quadratic equations?
2. Can you always solve quadratic equations with the quadratic formula?
3. What are some common methods to simplify equations involving squares?
4. How does completing the square compare to the quadratic formula?
5. What role does factoring play in solving higher-degree polynomials?

Tip:

When solving quadratic equations, always try factoring first if possible. It is quicker and avoids unnecessary calculations.