Solve the following inequations.
`sqrt(-x^(2)+6x-5) gt 8-2x`

Answer

To solve the inequality \( \sqrt{-x^2 + 6x - 5} > 8 - 2x \), we will follow these steps:

### Step 1: Set up the inequality
We start with the original inequality:
\[ \sqrt{-x^2 + 6x - 5} > 8 - 2x \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the inequality:
\[ -x^2 + 6x - 5 > (8 - 2x)^2 \]

### Step 3: Expand the right side
Now, we expand the right side:
\[ (8 - 2x)^2 = 64 - 32x + 4x^2 \]
So the inequality becomes:
\[ -x^2 + 6x - 5 > 64 - 32x + 4x^2 \]

### Step 4: Rearrange the inequality
Next, we move all terms to one side:
\[ -x^2 + 6x - 5 - 64 + 32x - 4x^2 > 0 \]
Combining like terms gives us:
\[ -5x^2 + 38x - 69 > 0 \]
Multiplying through by -1 (and flipping the inequality) gives:
\[ 5x^2 - 38x + 69 < 0 \]

### Step 5: Factor the quadratic
To factor \( 5x^2 - 38x + 69 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 5, b = -38, c = 69 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-38)^2 - 4 \cdot 5 \cdot 69 = 1444 - 1380 = 64 \]
Now substituting back into the quadratic formula:
\[ x = \frac{38 \pm \sqrt{64}}{10} = \frac{38 \pm 8}{10} \]
This gives us two solutions:
\[ x_1 = \frac{46}{10} = 4.6 \quad \text{and} \quad x_2 = \frac{30}{10} = 3 \]

### Step 6: Determine the intervals
The roots divide the number line into intervals. We need to test the sign of the quadratic in each interval:
1. \( (-\infty, 3) \)
2. \( (3, 4.6) \)
3. \( (4.6, \infty) \)

### Step 7: Test the intervals
- For \( x < 3 \) (e.g., \( x = 0 \)):
\[ 5(0)^2 - 38(0) + 69 = 69 > 0 \]
- For \( 3 < x < 4.6 \) (e.g., \( x = 4 \)):
\[ 5(4)^2 - 38(4) + 69 = 80 - 152 + 69 = -3 < 0 \]
- For \( x > 4.6 \) (e.g., \( x = 5 \)):
\[ 5(5)^2 - 38(5) + 69 = 125 - 190 + 69 = 4 > 0 \]

### Step 8: Conclusion
The quadratic \( 5x^2 - 38x + 69 < 0 \) is satisfied in the interval:
\[ (3, 4.6) \]

### Final Answer
Thus, the solution to the inequality is:
\[ x \in (3, 4.6) \]