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.

Learn how to solve the quadratic equation (x-2)^2 + 5(x-2) + 6 = 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, including substitution method and quadratic formula. Suitable for students in Grades 9-12.

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