We are given the equation:

\[
5n - 4 = \frac{n(5n - 3)}{2}
\]

To solve for \( n \), let's follow these steps:

### Step 1: Eliminate the fraction by multiplying both sides by 2.

\[
2(5n - 4) = n(5n - 3)
\]

This simplifies to:

\[
10n - 8 = n(5n - 3)
\]

### Step 2: Expand the right side of the equation.

\[
10n - 8 = 5n^2 - 3n
\]

### Step 3: Move all terms to one side to form a quadratic equation.

\[
0 = 5n^2 - 3n - 10n + 8
\]

Simplifying:

\[
0 = 5n^2 - 13n + 8
\]

### Step 4: Solve the quadratic equation using the quadratic formula.

The quadratic formula is given by:

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

For the quadratic equation \( 5n^2 - 13n + 8 = 0 \), we have:

- \( a = 5 \)
- \( b = -13 \)
- \( c = 8 \)

Plugging into the quadratic formula:

\[
n = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(5)(8)}}{2(5)}
\]

\[
n = \frac{13 \pm \sqrt{169 - 160}}{10}
\]

\[
n = \frac{13 \pm \sqrt{9}}{10}
\]

\[
n = \frac{13 \pm 3}{10}
\]

### Step 5: Find the two possible values of \( n \).

\[
n = \frac{13 + 3}{10} = \frac{16}{10} = 1.6
\]

or

\[
n = \frac{13 - 3}{10} = \frac{10}{10} = 1
\]

### Final Answer:
The two possible solutions are:

\[
n = 1.6 \quad \text{or} \quad n = 1
\]

Do you have any questions or would you like more details?

Here are 5 related questions to explore:

1. How do you determine if both solutions are valid in the original equation?
2. Can you solve the same quadratic using factoring methods?
3. How does the discriminant in the quadratic formula help us interpret the solution?
4. What are some real-life applications of solving quadratic equations like this?
5. How can we verify the solutions by substituting them back into the original equation?

**Tip**: Always check if the solutions you find satisfy the original equation by substituting them back in.

This detailed solution explains how to solve the quadratic equation (5n - 4) = n(5n - 3)/2 step by step. The solution involves simplifying, expanding, and applying the quadratic formula. Key concepts include algebra, solving quadratic equations, and discriminant analysis. Perfect for Grades 9-11 students.

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