Let's break down and solve the equation step by step.

The equation you provided is:

$\frac{3}{2} = \frac{11}{2x} \times \frac{x-2}{x}$

Step 1: Simplify the right-hand side.

The right-hand side is a product of two fractions:

$\frac{11}{2x} \times \frac{x-2}{x} = \frac{11(x-2)}{2x^2}$

So, the equation becomes:

$\frac{3}{2} = \frac{11(x-2)}{2x^2}$

Step 2: Eliminate the denominators.

Multiply both sides of the equation by $2x^2$ to remove the denominators:

$3 \cdot 2x^2 = 11(x - 2) \cdot 2$

This simplifies to:

$6x^2 = 22(x - 2)$

Step 3: Expand both sides.

Expand the right-hand side:

$6x^2 = 22x - 44$

Step 4: Rearrange the equation.

Move all terms to one side of the equation:

$6x^2 - 22x + 44 = 0$

Step 5: Simplify the equation.

Divide the entire equation by 2 to make it simpler:

$3x^2 - 11x + 22 = 0$

Step 6: Solve the quadratic equation.

Now we need to solve the quadratic equation $3x^2 - 11x + 22 = 0$. We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$, $b = -11$, and $c = 22$. Plug these values into the formula:

$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(3)(22)}}{2(3)}$ $x = \frac{11 \pm \sqrt{121 - 264}}{6}$ $x = \frac{11 \pm \sqrt{-143}}{6}$

Since $\sqrt{-143}$ is imaginary, the solutions are complex numbers:

$x = \frac{11 \pm i\sqrt{143}}{6}$

Final Answer:

The solutions to the equation are complex:

$x = \frac{11 + i\sqrt{143}}{6} \quad \text{or} \quad x = \frac{11 - i\sqrt{143}}{6}$

Would you like more details on any of these steps or have any questions?

Here are 5 related questions you might find interesting:

1. How do you solve quadratic equations with real solutions?
2. What are complex numbers and how are they used?
3. How can you simplify the quadratic formula to solve faster?
4. How do you check if a quadratic equation has real or complex solutions?
5. What are the properties of imaginary numbers?

Tip: Always check the discriminant ($b^2 - 4ac$) before solving a quadratic to determine if the solutions are real or complex.