Solve the following inequalities:
`(3x+4)/(x^(2)-3x+5)lt0`

Answer

To solve the inequality \(\frac{3x + 4}{x^2 - 3x + 5} < 0\), we will follow these steps:

### Step 1: Identify the numerator and denominator
We start with the inequality:
\[ \frac{3x + 4}{x^2 - 3x + 5} < 0 \]
The numerator is \(3x + 4\) and the denominator is \(x^2 - 3x + 5\).

### Step 2: Find the critical points
To find the critical points, we need to determine when the numerator and denominator are equal to zero.

**Numerator:**
Set \(3x + 4 = 0\):
\[ 3x = -4 \implies x = -\frac{4}{3} \]

**Denominator:**
Set \(x^2 - 3x + 5 = 0\) and calculate the discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 5 = 9 - 20 = -11 \]
Since the discriminant \(D < 0\), the quadratic \(x^2 - 3x + 5\) has no real roots and is always positive (as the leading coefficient is positive).

### Step 3: Analyze the sign of the expression
Since the denominator is always positive, the sign of the entire expression \(\frac{3x + 4}{x^2 - 3x + 5}\) will depend solely on the numerator \(3x + 4\).

**Numerator:**
We already found that \(3x + 4 = 0\) at \(x = -\frac{4}{3}\). We will analyze the sign of \(3x + 4\):

- For \(x < -\frac{4}{3}\), \(3x + 4 < 0\) (negative).
- For \(x > -\frac{4}{3}\), \(3x + 4 > 0\) (positive).

### Step 4: Determine the solution set
We need the expression to be less than zero:
\[ \frac{3x + 4}{x^2 - 3x + 5} < 0 \]
This occurs when \(3x + 4 < 0\), which is true for:
\[ x < -\frac{4}{3} \]

### Conclusion
The solution to the inequality is:
\[ x \in \left(-\infty, -\frac{4}{3}\right) \]