Solve each of the following quadratic equations:
`((4x-3)/(2x+1))-10((2x+1)/(4x-3))=3,xne(-1)/(2),(3)/(4)`

Answer

To solve the quadratic equation given by

\[ \frac{4x-3}{2x+1} - 10\left(\frac{2x+1}{4x-3}\right) = 3 \]

we will follow these steps:

### Step 1: Set up the equation
We start with the equation:

\[ \frac{4x-3}{2x+1} - 10\left(\frac{2x+1}{4x-3}\right) = 3 \]

### Step 2: Let \( y = \frac{4x-3}{2x+1} \)
We can simplify our equation by substituting \( y \):

\[ y - 10\left(\frac{1}{y}\right) = 3 \]

### Step 3: Multiply through by \( y \)
To eliminate the fraction, we multiply both sides by \( y \):

\[ y^2 - 10 = 3y \]

### Step 4: Rearrange the equation
Rearranging gives us a standard quadratic form:

\[ y^2 - 3y - 10 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor \( y^2 - 3y - 10 \):

\[ (y - 5)(y + 2) = 0 \]

### Step 6: Solve for \( y \)
Setting each factor to zero gives us:

1. \( y - 5 = 0 \) → \( y = 5 \)
2. \( y + 2 = 0 \) → \( y = -2 \)

### Step 7: Substitute back for \( x \)
Recall that \( y = \frac{4x-3}{2x+1} \). We will substitute back to find \( x \).

#### For \( y = 5 \):
\[ \frac{4x-3}{2x+1} = 5 \]

Cross-multiplying gives:

\[ 4x - 3 = 5(2x + 1) \]

Expanding and rearranging:

\[ 4x - 3 = 10x + 5 \]
\[ 4x - 10x = 5 + 3 \]
\[ -6x = 8 \]
\[ x = -\frac{4}{3} \]

#### For \( y = -2 \):
\[ \frac{4x-3}{2x+1} = -2 \]

Cross-multiplying gives:

\[ 4x - 3 = -2(2x + 1) \]

Expanding and rearranging:

\[ 4x - 3 = -4x - 2 \]
\[ 4x + 4x = -2 + 3 \]
\[ 8x = 1 \]
\[ x = \frac{1}{8} \]

### Final Solutions
The solutions to the quadratic equation are:

\[ x = -\frac{4}{3} \quad \text{and} \quad x = \frac{1}{8} \]