`2x^2+5x-3gt0`

Options

A. `(-oo,-3]uu(1/2,oo)`
B. `(-oo,-3[uu[2,oo]`
C. `R-[-3,1/2]`
D. `R-(-3,1/2)`

Answer

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

### Step 1: Factor the quadratic expression
We start with the quadratic inequality:
\[ 2x^2 + 5x - 3 > 0 \]
We will factor the quadratic expression. To do this, we can look for two numbers that multiply to \(2 \times -3 = -6\) and add to \(5\). The numbers \(6\) and \(-1\) satisfy this condition.

Rewriting the middle term:
\[ 2x^2 + 6x - x - 3 > 0 \]

### Step 2: Group the terms
Next, we group the terms:
\[ (2x^2 + 6x) + (-x - 3) > 0 \]

### Step 3: Factor by grouping
Now, we factor out the common factors from each group:
\[ 2x(x + 3) - 1(x + 3) > 0 \]
This can be factored further:
\[ (2x - 1)(x + 3) > 0 \]

### Step 4: Find the critical points
To solve the inequality, we need to find the critical points where the expression equals zero:
\[ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2} \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

### Step 5: Test intervals
Now we will test the intervals defined by the critical points \(x = -3\) and \(x = \frac{1}{2}\):
1. **Interval 1:** \( (-\infty, -3) \)
- Choose \(x = -4\): \((2(-4) - 1)(-4 + 3) = (-8 - 1)(-1) = 9 > 0\) (True)

2. **Interval 2:** \( (-3, \frac{1}{2}) \)
- Choose \(x = 0\): \((2(0) - 1)(0 + 3) = (-1)(3) = -3 < 0\) (False)

3. **Interval 3:** \( (\frac{1}{2}, \infty) \)
- Choose \(x = 1\): \((2(1) - 1)(1 + 3) = (2 - 1)(4) = 1 \cdot 4 = 4 > 0\) (True)

### Step 6: Write the solution
From our testing, the inequality \( (2x - 1)(x + 3) > 0 \) holds true in the intervals:
\[ (-\infty, -3) \quad \text{and} \quad \left(\frac{1}{2}, \infty\right) \]

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