Solve the inequation `(sqrt(6+x-x^(2)))/(2x+5) ge (sqrt(6+x-x^(2)))/(x+4)`.

Answer

To solve the inequation

\[ \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}, \]

we'll follow these steps:

### Step 1: Rewrite the Inequation
We start with the given inequation:

\[ \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}. \]

### Step 2: Cross Multiply
Assuming \(\sqrt{6+x-x^2} \geq 0\) (which we will verify later), we can cross-multiply:

\[ \sqrt{6+x-x^2} \cdot (x+4) \geq \sqrt{6+x-x^2} \cdot (2x+5). \]

### Step 3: Simplify the Inequation
We can factor out \(\sqrt{6+x-x^2}\):

\[ \sqrt{6+x-x^2} \left( x+4 - (2x+5) \right) \geq 0. \]

This simplifies to:

\[ \sqrt{6+x-x^2} \left( x + 4 - 2x - 5 \right) \geq 0, \]

which further simplifies to:

\[ \sqrt{6+x-x^2} \left( -x - 1 \right) \geq 0. \]

### Step 4: Analyze the Factors
Now we have two factors:

1. \(\sqrt{6+x-x^2}\)
2. \(-x - 1\)

#### Factor 1: \(\sqrt{6+x-x^2} \geq 0\)
For the square root to be defined and non-negative, we need:

\[ 6 + x - x^2 \geq 0. \]

This can be rearranged to:

\[ -x^2 + x + 6 \geq 0. \]

Factoring gives:

\[ -(x^2 - x - 6) \geq 0 \implies (x-3)(x+2) \leq 0. \]

The roots are \(x = -2\) and \(x = 3\). The solution to this inequality is:

\[ -2 \leq x \leq 3. \]

#### Factor 2: \(-x - 1 \geq 0\)
This simplifies to:

\[ -x \geq 1 \implies x \leq -1. \]

### Step 5: Combine the Results
Now we combine the results from both factors:

1. From \(\sqrt{6+x-x^2} \geq 0\): \(-2 \leq x \leq 3\).
2. From \(-x - 1 \geq 0\): \(x \leq -1\).

The overlapping solution is:

\[ -2 \leq x \leq -1. \]

### Final Answer
Thus, the solution to the inequation is:

\[ x \in [-2, -1]. \]