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Let's solve the given equation:

\[
\frac{6-x}{x^2 - 4} - 2 = \frac{x}{x+2}
\]

### Step 1: Simplifying the equation
The denominator \(x^2 - 4\) can be factored as:
\[
x^2 - 4 = (x-2)(x+2)
\]
So, the equation becomes:
\[
\frac{6-x}{(x-2)(x+2)} - 2 = \frac{x}{x+2}
\]

### Step 2: Finding a common denominator
The common denominator for all terms is \((x-2)(x+2)\). We can rewrite the equation as:
\[
\frac{6-x - 2(x^2-4)}{(x-2)(x+2)} = \frac{x(x-2)}{(x-2)(x+2)}
\]
Simplifying the numerator on the left side:
\[
6 - x - 2(x^2 - 4) = 6 - x - 2x^2 + 8 = -2x^2 - x + 14
\]
The equation now becomes:
\[
\frac{-2x^2 - x + 14}{(x-2)(x+2)} = \frac{x(x-2)}{(x-2)(x+2)}
\]

### Step 3: Setting the numerators equal to each other
Since the denominators are the same, we equate the numerators:
\[
-2x^2 - x + 14 = x(x-2)
\]
Simplifying the right-hand side:
\[
-2x^2 - x + 14 = x^2 - 2x
\]
Bringing all terms to the left side:
\[
-3x^2 + x + 14 = 0
\]

### Step 4: Solving the quadratic equation
We solve \( -3x^2 + x + 14 = 0 \) using the quadratic formula:
\[
x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
where \( A = -3 \), \( B = 1 \), and \( C = 14 \).

\[
x = \frac{-1 \pm \sqrt{1^2 - 4(-3)(14)}}{2(-3)}
\]
Simplifying:
\[
x = \frac{-1 \pm \sqrt{1 + 168}}{-6}
\]
\[
x = \frac{-1 \pm \sqrt{169}}{-6}
\]
\[
x = \frac{-1 \pm 13}{-6}
\]
So, the two solutions are:
\[
x = \frac{-1 + 13}{-6} = -2 \quad \text{and} \quad x = \frac{-1 - 13}{-6} = 2.33
\]

### Step 5: Checking for restrictions
The restrictions for \(x\) are \( x \neq -2 \) and \( x \neq 2 \).

Hence, the valid solution is \( x = 1 \).

Solve for x: \( \frac{6-x}{x^2 - 4} - 2 = \frac{x}{x+2} \), with restrictions \( x \neq -2 \) and \( x \neq 2 \).

Learn how to solve the rational equation \( \frac{6-x}{x^2 - 4} - 2 = \frac{x}{x+2} \) using algebraic manipulation and the quadratic formula. This problem covers concepts such as finding common denominators and applying the quadratic formula. Suitable for students in Grades 9-12.

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