Let's break down and solve the given equation step by step:

$(x^2 - 7x + 13) - (2 - 3)(x - 4) = 1$

Step 1: Simplify the terms

The expression $(2 - 3)$ simplifies to $-1$, so the equation becomes:

$(x^2 - 7x + 13) - (-1)(x - 4) = 1$

Step 2: Distribute the $-1$ on the second term:

Now distribute $-1$ across the terms in $(x - 4)$:

$(x^2 - 7x + 13) + (x - 4) = 1$

Step 3: Combine like terms

Next, combine the like terms from the left side of the equation:

$x^2 - 7x + x + 13 - 4 = 1$ $x^2 - 6x + 9 = 1$

Step 4: Set the equation to zero

To make it easier to solve, subtract 1 from both sides of the equation:

$x^2 - 6x + 9 - 1 = 0$ $x^2 - 6x + 8 = 0$

Step 5: Factor the quadratic

Now, factor the quadratic equation $x^2 - 6x + 8 = 0$. We look for two numbers that multiply to $8$ and add up to $-6$. These numbers are $-4$ and $-2$, so the factored form is:

$(x - 4)(x - 2) = 0$

Step 6: Solve for $x$

Set each factor equal to zero:

$x - 4 = 0 \quad \text{or} \quad x - 2 = 0$

$x = 4 \quad \text{or} \quad x = 2$

Final Answer:

The solutions to the equation are:

$x = 4 \quad \text{or} \quad x = 2$

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Would you like more details on any step or have further questions? Here are some related questions to explore:

1. How can we solve a quadratic equation using the quadratic formula?
2. What if the equation had no real solutions? How would we handle that?
3. What are the properties of quadratic equations and their graphs?
4. How do we factor quadratics with coefficients other than 1?
5. How can we use completing the square to solve quadratics?

Tip: Always check your factored solutions by plugging them back into the original equation to verify the correctness!