Let's solve the equation:

\[
5 + \frac{5}{x} - \frac{3}{2x} = \frac{3}{4x^2}
\]

### Step 1: Eliminate the denominators
To eliminate the fractions, multiply every term by \(4x^2\), the least common denominator (LCD):

\[
4x^2 \left(5 + \frac{5}{x} - \frac{3}{2x}\right) = 4x^2 \cdot \frac{3}{4x^2}
\]

Distribute \(4x^2\) to each term:

\[
4x^2 \cdot 5 + 4x^2 \cdot \frac{5}{x} - 4x^2 \cdot \frac{3}{2x} = 3
\]

Simplify each term:

\[
20x^2 + 20x - 6x = 3
\]

Combine like terms:

\[
20x^2 + 14x = 3
\]

### Step 2: Rearrange the equation into standard quadratic form
Subtract 3 from both sides:

\[
20x^2 + 14x - 3 = 0
\]

### Step 3: Solve the quadratic equation
You can solve this using the quadratic formula:

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

Here, \(a = 20\), \(b = 14\), and \(c = -3\):

\[
x = \frac{-14 \pm \sqrt{14^2 - 4(20)(-3)}}{2(20)}
\]
\[
x = \frac{-14 \pm \sqrt{196 + 240}}{40}
\]
\[
x = \frac{-14 \pm \sqrt{436}}{40}
\]

### Step 4: Simplify the square root
\[
\sqrt{436} \approx 20.88
\]
\[
x = \frac{-14 \pm 20.88}{40}
\]

So, the two possible solutions for \(x\) are:

\[
x_1 = \frac{-14 + 20.88}{40} \approx \frac{6.88}{40} \approx 0.172
\]

\[
x_2 = \frac{-14 - 20.88}{40} \approx \frac{-34.88}{40} \approx -0.872
\]

### Solutions:
\[
x_1 \approx 0.172, \quad x_2 \approx -0.872
\]

Would you like further details on any part of this solution or have any questions?

### Related Questions
1. How can the discriminant of a quadratic equation help determine the number of real solutions?
2. What are the general steps to solve any quadratic equation?
3. Can you explain how to factor a quadratic equation when possible?
4. How does completing the square work in solving quadratic equations?
5. What are some real-life applications of quadratic equations?

### Tip
Always check for extraneous solutions when dealing with equations involving fractions, especially when multiplying both sides by variables.

Learn how to solve the equation 5 + 5/x - 3/(2x) = 3/(4x²) step-by-step with detailed explanations. This problem involves solving equations with fractions and quadratic elements. Suitable for high school students.

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