`(sqrt(8+2x-x^2)>6-3x)`

Explanation

To solve the inequality \( \sqrt{8 + 2x - x^2} > 6 - 3x \), we will follow these steps:

### Step 1: Determine the domain of the square root
The expression under the square root must be non-negative:
\[ 8 + 2x - x^2 \geq 0 \]
Rearranging gives:
\[ -x^2 + 2x + 8 \geq 0 \]
This can be rewritten as:
\[ -x^2 + 2x + 8 = -(x^2 - 2x - 8) \]
To find the roots, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = -8 \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]
Calculating the roots:
\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-4}{2} = -2 \]
Thus, the roots are \( x = -2 \) and \( x = 4 \).

### Step 2: Analyze the sign of the quadratic
The quadratic \( -(x - 4)(x + 2) \) is a downward-opening parabola. The expression is non-negative between the roots:
\[ -2 \leq x \leq 4 \]

### Step 3: Solve the inequality
Now we will consider the inequality:
\[ \sqrt{8 + 2x - x^2} > 6 - 3x \]
Squaring both sides (valid since both sides are non-negative in the domain):
\[ 8 + 2x - x^2 > (6 - 3x)^2 \]
Expanding the right side:
\[ 8 + 2x - x^2 > 36 - 36x + 9x^2 \]
Rearranging gives:
\[ 0 > 10x^2 - 38x + 28 \]
This simplifies to:
\[ 10x^2 - 38x + 28 < 0 \]

### Step 4: Factor the quadratic
To factor \( 10x^2 - 38x + 28 \), we can use the quadratic formula again:
\[ x = \frac{-(-38) \pm \sqrt{(-38)^2 - 4 \cdot 10 \cdot 28}}{2 \cdot 10} = \frac{38 \pm \sqrt{1444 - 1120}}{20} = \frac{38 \pm \sqrt{324}}{20} = \frac{38 \pm 18}{20} \]
Calculating the roots:
\[ x = \frac{56}{20} = \frac{14}{5} \quad \text{and} \quad x = \frac{20}{20} = 1 \]

### Step 5: Determine the intervals
The quadratic \( 10x^2 - 38x + 28 \) is negative between its roots:
\[ 1 < x < \frac{14}{5} \]

### Step 6: Find the intersection with the domain
We need to find the intersection of the intervals:
1. From the square root: \( -2 \leq x \leq 4 \)
2. From the quadratic: \( 1 < x < \frac{14}{5} \)

The intersection is:
\[ 1 < x < \frac{14}{5} \]

### Step 7: Final solution
Thus, the solution to the inequality \( \sqrt{8 + 2x - x^2} > 6 - 3x \) is:
\[ x \in (1, \frac{14}{5}) \]