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You uploaded an equation image. The equation is:

\[
(x - 2)^2 - (2x + 3)^2 = (x - 1)^2 - 6
\]

Let me solve this equation step by step.

### Step 1: Expand both sides

First, expand all the squared terms on both sides of the equation.

\[
(x - 2)^2 = x^2 - 4x + 4
\]
\[
(2x + 3)^2 = 4x^2 + 12x + 9
\]
\[
(x - 1)^2 = x^2 - 2x + 1
\]

Now substitute these into the original equation:

\[
(x^2 - 4x + 4) - (4x^2 + 12x + 9) = (x^2 - 2x + 1) - 6
\]

### Step 2: Simplify both sides

Simplify both the left-hand side (LHS) and right-hand side (RHS).

#### Left-hand side:
\[
(x^2 - 4x + 4) - (4x^2 + 12x + 9) = x^2 - 4x + 4 - 4x^2 - 12x - 9
\]
\[
= -3x^2 - 16x - 5
\]

#### Right-hand side:
\[
(x^2 - 2x + 1) - 6 = x^2 - 2x - 5
\]

### Step 3: Set the simplified equation

Now set the simplified forms equal to each other:

\[
-3x^2 - 16x - 5 = x^2 - 2x - 5
\]

### Step 4: Move all terms to one side

Move all terms to one side to set the equation to 0:

\[
-3x^2 - 16x - 5 - x^2 + 2x + 5 = 0
\]
\[
-4x^2 - 14x = 0
\]

### Step 5: Factor the equation

Factor out the common term:

\[
-2x(2x + 7) = 0
\]

### Step 6: Solve for \(x\)

The solutions are:

\[
x = 0 \quad \text{or} \quad 2x + 7 = 0
\]
\[
x = 0 \quad \text{or} \quad x = -\frac{7}{2}
\]

Thus, the solutions to the equation are:

\[
x = 0 \quad \text{or} \quad x = -\frac{7}{2}
\]

---

Do you need any further details or have any questions?

### Related Questions:
1. What are the steps to solve quadratic equations by factoring?
2. How can you expand squared binomials like \( (a + b)^2 \)?
3. What is the difference between factoring and expanding polynomials?
4. How do you handle equations with multiple squared terms?
5. Can you solve the same equation using the quadratic formula?

### Tip:
Always simplify each side of an equation completely before moving terms around, as this reduces mistakes.

This tutorial explains how to solve the equation (x - 2)^2 - (2x + 3)^2 = (x - 1)^2 - 6 by expanding binomials and factoring quadratic expressions. It covers the algebraic techniques needed for students in Grades 9-10.

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