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solve the inequality by greater or equal to sign 

The inequality provided is:

\[
2x^2 \geq -7x + 15
\]

To solve this inequality, let's rearrange it to set it to zero:

\[
2x^2 + 7x - 15 \geq 0
\]

### Step 1: Solve the related equation
First, solve the equation \(2x^2 + 7x - 15 = 0\) to find the boundary points. We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 2\), \(b = 7\), and \(c = -15\).

\[
x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2}
\]
\[
x = \frac{-7 \pm \sqrt{49 + 120}}{4}
\]
\[
x = \frac{-7 \pm \sqrt{169}}{4}
\]
\[
x = \frac{-7 \pm 13}{4}
\]

This gives two solutions:

1. \( x = \frac{-7 + 13}{4} = \frac{6}{4} = 1.5 \)
2. \( x = \frac{-7 - 13}{4} = \frac{-20}{4} = -5 \)

So, the boundary points are \( x = -5 \) and \( x = 1.5 \).

### Step 2: Test intervals
Now, we test the intervals determined by these boundary points: \( (-\infty, -5) \), \( (-5, 1.5) \), and \( (1.5, \infty) \).

1. **For \(x \in (-\infty, -5)\):** Pick \(x = -6\).
   - \(2(-6)^2 + 7(-6) - 15 = 72 - 42 - 15 = 15\), which is positive.

2. **For \(x \in (-5, 1.5)\):** Pick \(x = 0\).
   - \(2(0)^2 + 7(0) - 15 = -15\), which is negative.

3. **For \(x \in (1.5, \infty)\):** Pick \(x = 2\).
   - \(2(2)^2 + 7(2) - 15 = 8 + 14 - 15 = 7\), which is positive.

### Step 3: Write the solution
The inequality \(2x^2 + 7x - 15 \geq 0\) holds in the intervals where the expression is positive or zero, which are:

\[
x \in (-\infty, -5] \cup [1.5, \infty)
\]

### Final Answer:
The solution to the inequality is:

\[
x \in (-\infty, -5] \cup [1.5, \infty)
\]

Would you like a more detailed explanation or have any questions?

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Here are some related questions to consider:

1. How do we determine which intervals to test in an inequality?
2. What is the purpose of using the quadratic formula in solving inequalities?
3. How can we verify if the boundary points satisfy the inequality?
4. Can we solve quadratic inequalities by graphing instead of algebraically?
5. How do we handle cases where there are complex solutions?

**Tip:** When solving inequalities, always check each interval around the critical points to ensure the inequality holds as expected.

Solve the inequality by greater or equal to sign: 2x^2 - 7x + 15 ≥ 0

Learn how to solve the quadratic inequality 2x^2 - 7x + 15 ≥ 0 by finding boundary points and testing intervals. This solution uses the quadratic formula and interval testing, covering essential algebraic concepts suitable for Grades 9-11.

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