To solve the equation $(x-2)^2 + 5(x-2) + 6 = 0$, we can start by making a substitution to simplify it.

Let $y = x - 2$. This gives us:

$y^2 + 5y + 6 = 0$

Now, we need to solve this quadratic equation. The standard form of a quadratic equation is $ay^2 + by + c = 0$. In this case, $a = 1$, $b = 5$, and $c = 6$.

We can use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

$y = \frac{-5 \pm \sqrt{25 - 24}}{2}$

$y = \frac{-5 \pm \sqrt{1}}{2}$

$y = \frac{-5 \pm 1}{2}$

So, we have two solutions for $y$:

$y = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$

$y = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$

Since $y = x - 2$, we can now substitute back to find the values of $x$:

For $y = -2$:

$x - 2 = -2$

$x = 0$

For $y = -3$:

$x - 2 = -3$

$x = -1$

Therefore, the solutions to the original equation are $x = 0$ and $x = -1$.

Would you like more details or have any questions?

Here are 5 more questions you can ask:

1. How do you solve a system of linear equations?
2. Can you explain the process of factoring quadratic equations?
3. What is the derivative of a function, and how is it calculated?
4. How do you find the integral of a function?
5. Can you explain the properties of logarithms?

Tip: When solving quadratic equations, always check if they can be factored easily before using the quadratic formula, as factoring can be a quicker method.