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Let's solve the problem step-by-step.

### Given equation:
\[
24 = x^2 - 4x + 3
\]

### Part 1: Completing the square

We want to rewrite the quadratic equation by completing the square. Follow these steps:

1. Move the constant term (24) to the other side:
   \[
   x^2 - 4x + 3 = 24 \quad \Rightarrow \quad x^2 - 4x + 3 - 24 = 0 \quad \Rightarrow \quad x^2 - 4x - 21 = 0
   \]
2. To complete the square, take the coefficient of \(x\) (which is \(-4\)), divide it by 2, and square it:
   \[
   \left(\frac{-4}{2}\right)^2 = 4
   \]
   Add and subtract this value inside the equation:
   \[
   x^2 - 4x + 4 - 4 - 21 = 0 \quad \Rightarrow \quad (x - 2)^2 - 25 = 0
   \]
3. The equation now becomes:
   \[
   (x - 2)^2 = 25
   \]

So, we have completed the square:
\[
(x - 2)^2 = 25
\]

### Part 2: Solving the equation

Now, solve for \(x\) from the equation \((x - 2)^2 = 25\):

1. Take the square root of both sides:
   \[
   x - 2 = \pm \sqrt{25} = \pm 5
   \]
2. Solve for \(x\):
   \[
   x = 2 \pm 5
   \]

This gives two solutions:
\[
x = 2 + 5 = 7 \quad \text{or} \quad x = 2 - 5 = -3
\]

### Correct Answer:
The solutions are \(x = 7\) and \(x = -3\), which matches option A: \(x = 2 \pm 5\).

Would you like further clarification on any step?

Here are 5 related questions to explore:
1. How do you complete the square for a quadratic with a non-1 coefficient for \(x^2\)?
2. How do we factor quadratics without completing the square?
3. What is the quadratic formula, and how does it compare to completing the square?
4. How can you verify solutions of quadratic equations by substitution?
5. How does completing the square help in solving real-world problems?

**Tip:** Completing the square is particularly useful in deriving the quadratic formula and for working with circles and ellipses in geometry!

Consider the equation: 24 = x^2 - 4x + 3. Rewrite the equation by completing the square and find the solutions to the equation.

Learn how to solve the quadratic equation x^2 - 4x + 3 = 24 by completing the square. This step-by-step solution covers key algebraic concepts and is suitable for students in Grades 8-10.

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