The set of real values of x satisfying the inequality `|x^(2) + x -6| lt 6`, is

Options

A. `(-4, 3)`
B. `(-3, 2)`
C. `(-4, -3)cup(2, 3)`
D. `(-4, -1) cup (0, 3)`

Explanation

To solve the inequality \( |x^2 + x - 6| < 6 \), we will break it down into two separate inequalities:

1. \( x^2 + x - 6 < 6 \)
2. \( x^2 + x - 6 > -6 \)

### Step 1: Solve the first inequality \( x^2 + x - 6 < 6 \)

Rearranging the inequality gives:
\[ x^2 + x - 6 - 6 < 0 \implies x^2 + x - 12 < 0 \]

Next, we will factor the quadratic expression:
\[ x^2 + x - 12 = (x + 4)(x - 3) \]

Now we need to find the intervals where \( (x + 4)(x - 3) < 0 \).

### Step 2: Determine the critical points

The critical points are found by setting the factors to zero:
\[ x + 4 = 0 \implies x = -4 \]
\[ x - 3 = 0 \implies x = 3 \]

### Step 3: Test intervals

We will test the intervals determined by the critical points: \( (-\infty, -4) \), \( (-4, 3) \), and \( (3, \infty) \).

- For \( x < -4 \) (e.g., \( x = -5 \)):
\[ (-5 + 4)(-5 - 3) = (-1)(-8) = 8 > 0 \]

- For \( -4 < x < 3 \) (e.g., \( x = 0 \)):
\[ (0 + 4)(0 - 3) = (4)(-3) = -12 < 0 \]

- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ (4 + 4)(4 - 3) = (8)(1) = 8 > 0 \]

Thus, the solution for the first inequality is:
\[ -4 < x < 3 \]

### Step 4: Solve the second inequality \( x^2 + x - 6 > -6 \)

Rearranging gives:
\[ x^2 + x - 6 + 6 > 0 \implies x^2 + x > 0 \]

Factoring:
\[ x(x + 1) > 0 \]

### Step 5: Determine the critical points

The critical points are:
\[ x = 0 \quad \text{and} \quad x + 1 = 0 \implies x = -1 \]

### Step 6: Test intervals

We will test the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 0) \), and \( (0, \infty) \).

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2)(-2 + 1) = (-2)(-1) = 2 > 0 \]

- For \( -1 < x < 0 \) (e.g., \( x = -0.5 \)):
\[ (-0.5)(-0.5 + 1) = (-0.5)(0.5) = -0.25 < 0 \]

- For \( x > 0 \) (e.g., \( x = 1 \)):
\[ (1)(1 + 1) = (1)(2) = 2 > 0 \]

Thus, the solution for the second inequality is:
\[ x < -1 \quad \text{or} \quad x > 0 \]

### Step 7: Combine the solutions

Now we combine the solutions from both inequalities:
1. From \( -4 < x < 3 \)
2. From \( x < -1 \) or \( x > 0 \)

The combined solution is:
- From the first inequality: \( -4 < x < 3 \)
- From the second inequality: \( x < -1 \) gives \( -4 < x < -1 \)
- From the second inequality: \( x > 0 \) gives \( 0 < x < 3 \)

Thus, the final solution is:
\[ x \in (-4, -1) \cup (0, 3) \]

### Final Answer:
The set of real values of \( x \) satisfying the inequality \( |x^2 + x - 6| < 6 \) is:
\[ (-4, -1) \cup (0, 3) \]