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The given limit problem is:

\[
\lim_{x \to 2} \frac{\sqrt{3x^2 + 4} - 2\sqrt{8x - 3x^2}}{x^2 - 2x}
\]

### Step 1: Check Direct Substitution
First, let's try substituting \(x = 2\) directly into the expression:

- Numerator:
  \[
  \sqrt{3(2)^2 + 4} - 2\sqrt{8(2) - 3(2)^2} = \sqrt{3(4) + 4} - 2\sqrt{16 - 12}
  \]
  \[
  = \sqrt{12 + 4} - 2\sqrt{4} = \sqrt{16} - 2(2) = 4 - 4 = 0
  \]

- Denominator:
  \[
  (2)^2 - 2(2) = 4 - 4 = 0
  \]

This results in a **0/0 indeterminate form**, so we must use algebraic techniques to simplify the expression.

### Step 2: Rationalize the Numerator
We can rationalize the numerator by multiplying the expression by the conjugate of the numerator:

\[
\frac{\sqrt{3x^2 + 4} - 2\sqrt{8x - 3x^2}}{x^2 - 2x} \cdot \frac{\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2}}{\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2}}
\]

This results in:

\[
\frac{(\sqrt{3x^2 + 4})^2 - (2\sqrt{8x - 3x^2})^2}{(x^2 - 2x)(\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2})}
\]

Now simplify the numerator:

- \((\sqrt{3x^2 + 4})^2 = 3x^2 + 4\)
- \((2\sqrt{8x - 3x^2})^2 = 4(8x - 3x^2) = 32x - 12x^2\)

So, the numerator becomes:

\[
(3x^2 + 4) - (32x - 12x^2) = 3x^2 + 4 - 32x + 12x^2 = 15x^2 - 32x + 4
\]

Thus, the expression is now:

\[
\frac{15x^2 - 32x + 4}{(x^2 - 2x)(\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2})}
\]

### Step 3: Factor the Denominator
The denominator \(x^2 - 2x\) can be factored as:

\[
x^2 - 2x = x(x - 2)
\]

So the expression becomes:

\[
\frac{15x^2 - 32x + 4}{x(x - 2)(\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2})}
\]

### Step 4: Factor the Numerator
Now, let's factor the quadratic \(15x^2 - 32x + 4\). Using the quadratic formula:

\[
x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(15)(4)}}{2(15)}
\]
\[
x = \frac{32 \pm \sqrt{1024 - 240}}{30} = \frac{32 \pm \sqrt{784}}{30} = \frac{32 \pm 28}{30}
\]

So, the roots are:

\[
x = \frac{32 + 28}{30} = \frac{60}{30} = 2 \quad \text{and} \quad x = \frac{32 - 28}{30} = \frac{4}{30} = \frac{2}{15}
\]

Thus, we can factor the numerator as:

\[
15x^2 - 32x + 4 = 15(x - 2)\left(x - \frac{2}{15}\right)
\]

### Step 5: Simplify the Expression
Now the limit becomes:

\[
\lim_{x \to 2} \frac{15(x - 2)\left(x - \frac{2}{15}\right)}{x(x - 2)(\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2})}
\]

Cancel the common factor of \(x - 2\):

\[
\lim_{x \to 2} \frac{15\left(x - \frac{2}{15}\right)}{x(\sqrt{3x^2 + 4} + 2\sqrt{8x - 3x^2})}
\]

### Step 6: Substitute \(x = 2\)
Now substitute \(x = 2\) into the remaining expression:

- Numerator:
  \[
  15\left(2 - \frac{2}{15}\right) = 15\left(\frac{30}{15} - \frac{2}{15}\right) = 15\left(\frac{28}{15}\right) = 28
  \]

- Denominator:
  \[
  2\left(\sqrt{3(2)^2 + 4} + 2\sqrt{8(2) - 3(2)^2}\right) = 2\left(\sqrt{12 + 4} + 2\sqrt{16 - 12}\right)
  \]
  \[
  = 2\left(\sqrt{16} + 2\sqrt{4}\right) = 2(4 + 2(2)) = 2(4 + 4) = 2(8) = 16
  \]

Thus, the limit is:

\[
\frac{28}{16} = \frac{7}{4}
\]

### Final Answer:
The limit is:

\[
\boxed{\frac{7}{4}}
\]

---

Would you like further clarification on any of the steps, or additional related questions? Here are some suggestions:

1. How do you factor complex quadratic expressions?
2. What are some alternative ways to simplify limits with square roots?
3. How does the quadratic formula help in solving limits?
4. Why is it important to rationalize both numerators and denominators in limits?
5. When is factoring the best technique for simplifying limits?

**Tip**: Always look for common factors like \(x - a\) in limits approaching \(x = a\), which can often cancel out indeterminate forms.

lim(x -> 2) sqrt(3x^2 + 4) - 2 sqrt(8x - 3x^2) / (x^2 - 2x)

This detailed step-by-step solution guides you through solving the limit of sqrt(3x^2 + 4) - 2sqrt(8x - 3x^2) over x^2 - 2x as x approaches 2. Using algebraic techniques like rationalizing the numerator and factoring, the solution simplifies the expression to give the final answer 7/4. Suitable for Grades 10-12.

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