We are given the quadratic inequality:

\[
3x^2 - 15x + 18 > 0
\]

### Step 1: Solve the corresponding equation
First, solve the equation \( 3x^2 - 15x + 18 = 0 \).

Divide through by 3 to simplify:

\[
x^2 - 5x + 6 = 0
\]

Now, factor the quadratic:

\[
x^2 - 5x + 6 = (x - 2)(x - 3) = 0
\]

The solutions to this equation are:

\[
x = 2 \quad \text{or} \quad x = 3
\]

### Step 2: Analyze the sign of the quadratic
Now, we analyze where the expression \( 3x^2 - 15x + 18 \) (or equivalently \( (x - 2)(x - 3) \)) is greater than 0.

The roots \( x = 2 \) and \( x = 3 \) divide the real number line into three intervals:

1. \( (-\infty, 2) \)
2. \( (2, 3) \)
3. \( (3, \infty) \)

### Step 3: Test the intervals
For each interval, pick a test point and determine the sign of \( (x - 2)(x - 3) \).

- For \( x \in (-\infty, 2) \), choose \( x = 0 \):  
  \((0 - 2)(0 - 3) = (-2)(-3) = 6 > 0\)

- For \( x \in (2, 3) \), choose \( x = 2.5 \):  
  \((2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0\)

- For \( x \in (3, \infty) \), choose \( x = 4 \):  
  \((4 - 2)(4 - 3) = (2)(1) = 2 > 0\)

### Step 4: Conclusion
The expression is positive on the intervals \( (-\infty, 2) \) and \( (3, \infty) \). Therefore, the solution to the inequality is:

\[
x \in (-\infty, 2) \cup (3, \infty)
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you solve quadratic inequalities with different methods?
2. Can a quadratic inequality have no real solution?
3. What is the graphical interpretation of this quadratic inequality?
4. How can you apply the quadratic formula to solve inequalities?
5. How do quadratic inequalities change when coefficients are negative?

**Tip**: Always check the sign of each factor in the intervals when solving inequalities by factoring.

Learn how to solve the quadratic inequality 3x^2 - 15x + 18 > 0 through factoring and sign analysis. This problem demonstrates key algebraic methods, such as solving quadratic equations by factorization and analyzing intervals. Suitable for students in Grades 9-12.

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