Solve `2x^(2) + 5x + 3 gt 0`

Answer

To solve the inequality \(2x^2 + 5x + 3 > 0\), we will follow these steps:

### Step 1: Identify the quadratic expression
We start with the inequality:
\[ 2x^2 + 5x + 3 > 0 \]

### Step 2: Factor the quadratic expression
We need to factor the quadratic expression \(2x^2 + 5x + 3\). We can rewrite it as:
\[ 2x^2 + 2x + 3x + 3 > 0 \]
Now, we can group the terms:
\[ (2x^2 + 2x) + (3x + 3) > 0 \]
Factoring out the common terms, we have:
\[ 2x(x + 1) + 3(x + 1) > 0 \]
Now we can factor by grouping:
\[ (2x + 3)(x + 1) > 0 \]

### Step 3: Find the roots of the equation
To find the intervals where the product is positive, we first find the roots of the equation \( (2x + 3)(x + 1) = 0 \):
1. Set \(2x + 3 = 0\):
\[ 2x = -3 \implies x = -\frac{3}{2} \]
2. Set \(x + 1 = 0\):
\[ x = -1 \]

### Step 4: Determine the intervals
The roots divide the number line into intervals. The critical points are \(x = -\frac{3}{2}\) and \(x = -1\). The intervals to test are:
1. \( (-\infty, -\frac{3}{2}) \)
2. \( (-\frac{3}{2}, -1) \)
3. \( (-1, \infty) \)

### Step 5: Test the intervals
We will choose test points from each interval to determine where the product is positive.

1. For the interval \( (-\infty, -\frac{3}{2}) \), choose \(x = -2\):
\[ (2(-2) + 3)(-2 + 1) = (-4 + 3)(-1) = (-1)(-1) = 1 > 0 \]
So, this interval satisfies the inequality.

2. For the interval \( (-\frac{3}{2}, -1) \), choose \(x = -1.25\):
\[ (2(-1.25) + 3)(-1.25 + 1) = (-2.5 + 3)(-0.25) = (0.5)(-0.25) = -0.125 < 0 \]
This interval does not satisfy the inequality.

3. For the interval \( (-1, \infty) \), choose \(x = 0\):
\[ (2(0) + 3)(0 + 1) = (3)(1) = 3 > 0 \]
This interval satisfies the inequality.

### Step 6: Combine the results
The inequality \(2x^2 + 5x + 3 > 0\) holds in the intervals:
\[ (-\infty, -\frac{3}{2}) \cup (-1, \infty) \]

### Final Answer
Thus, the solution to the inequality \(2x^2 + 5x + 3 > 0\) is:
\[ x \in (-\infty, -\frac{3}{2}) \cup (-1, \infty) \]