Solve `(2x+3)/(x^(2)+x-12) lt (1)/(2)`.

Explanation

To solve the inequality \(\frac{2x+3}{x^2+x-12} < \frac{1}{2}\), we will follow these steps:

### Step 1: Rearranging the Inequality
First, we will move \(\frac{1}{2}\) to the left side of the inequality:

\[ \frac{2x+3}{x^2+x-12} - \frac{1}{2} < 0 \]

### Step 2: Finding a Common Denominator
To combine the fractions, we need a common denominator. The common denominator will be \(2(x^2+x-12)\):

\[ \frac{2(2x+3) - (x^2+x-12)}{2(x^2+x-12)} < 0 \]

### Step 3: Simplifying the Numerator
Now, we simplify the numerator:

\[ 2(2x+3) = 4x + 6 \]
\[ -(x^2+x-12) = -x^2 - x + 12 \]
Combining these:

\[ 4x + 6 - x^2 - x + 12 = -x^2 + 3x + 18 \]

So, the inequality now looks like:

\[ \frac{-x^2 + 3x + 18}{2(x^2+x-12)} < 0 \]

### Step 4: Factoring the Numerator and Denominator
Next, we will factor the numerator and denominator.

**Numerator:**
To factor \(-x^2 + 3x + 18\), we can rewrite it as:

\[ -(x^2 - 3x - 18) \]

Factoring \(x^2 - 3x - 18\):

\[ x^2 - 3x - 18 = (x - 6)(x + 3) \]

Thus, the numerator becomes:

\[ -(x - 6)(x + 3) \]

**Denominator:**
Now, we factor the denominator \(x^2 + x - 12\):

\[ x^2 + x - 12 = (x - 3)(x + 4) \]

### Step 5: Rewrite the Inequality
Now we can rewrite the inequality:

\[ \frac{-(x - 6)(x + 3)}{2(x - 3)(x + 4)} < 0 \]

### Step 6: Analyzing the Sign of the Expression
To find the intervals where this expression is negative, we will find the critical points by setting the numerator and denominator to zero:

- **Numerator:** \( -(x - 6)(x + 3) = 0 \) gives \( x = 6 \) and \( x = -3 \).
- **Denominator:** \( 2(x - 3)(x + 4) = 0 \) gives \( x = 3 \) and \( x = -4 \).

The critical points are \( x = -4, -3, 3, 6 \).

### Step 7: Test Intervals
We will test the sign of the expression in the intervals defined by these critical points:

1. \( (-\infty, -4) \)
2. \( (-4, -3) \)
3. \( (-3, 3) \)
4. \( (3, 6) \)
5. \( (6, \infty) \)

**Testing:**
- For \( x < -4 \) (e.g., \( x = -5 \)): Positive
- For \( -4 < x < -3 \) (e.g., \( x = -3.5 \)): Negative
- For \( -3 < x < 3 \) (e.g., \( x = 0 \)): Positive
- For \( 3 < x < 6 \) (e.g., \( x = 4 \)): Negative
- For \( x > 6 \) (e.g., \( x = 7 \)): Positive

### Step 8: Conclusion
The expression is negative in the intervals \( (-4, -3) \) and \( (3, 6) \). Therefore, the solution to the inequality is:

\[ x \in (-4, -3) \cup (3, 6) \]