Find the values of x for which the inequality `(8x^(2)+16x-51)/((2x-3)(x+4))gt3` holds.

Answer

To solve the inequality \((8x^{2}+16x-51)/((2x-3)(x+4)) > 3\), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by moving \(3\) to the left side:
\[ \frac{8x^{2}+16x-51}{(2x-3)(x+4)} - 3 > 0 \]

### Step 2: Finding a Common Denominator
To combine the terms, we need a common denominator:
\[ \frac{8x^{2}+16x-51 - 3(2x-3)(x+4)}{(2x-3)(x+4)} > 0 \]

### Step 3: Expanding the Terms
Now, we expand \(3(2x-3)(x+4)\):
\[ 3(2x-3)(x+4) = 3(2x^2 + 8x - 3x - 12) = 3(2x^2 + 5x - 12) = 6x^2 + 15x - 36 \]

### Step 4: Simplifying the Numerator
Now, we substitute this back into the inequality:
\[ 8x^{2}+16x-51 - (6x^{2}+15x-36) > 0 \]
This simplifies to:
\[ (8x^{2} - 6x^{2}) + (16x - 15x) + (-51 + 36) > 0 \]
\[ 2x^{2} + x - 15 > 0 \]

### Step 5: Factoring the Quadratic
Next, we factor the quadratic \(2x^{2} + x - 15\):
\[ 2x^{2} + 6x - 5x - 15 = (2x - 5)(x + 3) \]
Thus, we have:
\[ (2x - 5)(x + 3) > 0 \]

### Step 6: Finding Critical Points
The critical points are found by setting each factor to zero:
\[ 2x - 5 = 0 \Rightarrow x = \frac{5}{2} \]
\[ x + 3 = 0 \Rightarrow x = -3 \]

### Step 7: Testing Intervals
We will test the intervals defined by the critical points \(-3\) and \(\frac{5}{2}\):
1. **Interval \((-∞, -3)\)**: Choose \(x = -4\):
\((2(-4) - 5)(-4 + 3) = (-8 - 5)(-1) = 13 > 0\) (True)

2. **Interval \((-3, \frac{5}{2})\)**: Choose \(x = 0\):
\((2(0) - 5)(0 + 3) = (-5)(3) = -15 < 0\) (False)

3. **Interval \((\frac{5}{2}, ∞)\)**: Choose \(x = 3\):
\((2(3) - 5)(3 + 3) = (6 - 5)(6) = 1 \cdot 6 = 6 > 0\) (True)

### Step 8: Conclusion
The solution to the inequality is:
\[ x \in (-\infty, -3) \cup \left(\frac{5}{2}, \infty\right) \]