We are tasked with solving the inequality:

$2x^2 + 6x - 9 < 0$

Step 1: Simplify the inequality.

We start by dividing the entire inequality by 2 to simplify it:

$x^2 + 3x - \frac{9}{2} < 0$

Step 2: Find the roots of the equation.

To find when $x^2 + 3x - \frac{9}{2} = 0$, we can use the quadratic formula:

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

Here, $a = 1$, $b = 3$, and $c = -\frac{9}{2}$.

$x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)\left(-\frac{9}{2}\right)}}{2(1)}$ $x = \frac{-3 \pm \sqrt{9 + 18}}{2}$ $x = \frac{-3 \pm \sqrt{27}}{2}$ $x = \frac{-3 \pm 3\sqrt{3}}{2}$ So, the roots are:

$x = \frac{-3 + 3\sqrt{3}}{2} \quad \text{or} \quad x = \frac{-3 - 3\sqrt{3}}{2}$

Step 3: Analyze the sign of the quadratic.

The quadratic $x^2 + 3x - \frac{9}{2}$ changes signs at these two roots. We can test the sign of the expression in the intervals divided by these roots:

1. $x < \frac{-3 - 3\sqrt{3}}{2}$
2. $\frac{-3 - 3\sqrt{3}}{2} < x < \frac{-3 + 3\sqrt{3}}{2}$
3. $x > \frac{-3 + 3\sqrt{3}}{2}$

Since the parabola opens upwards (positive $a = 1$), the expression will be negative between the roots and positive outside of them.

Step 4: Write the solution.

The solution to the inequality is:

$\frac{-3 - 3\sqrt{3}}{2} < x < \frac{-3 + 3\sqrt{3}}{2}$

Would you like further details or clarification?

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Here are 5 related questions to expand your understanding:

1. How do we solve quadratic inequalities with more complex roots?
2. Can you explain how the sign of the quadratic expression changes?
3. How does the quadratic formula work for different types of roots (real vs complex)?
4. Why does the direction of the inequality not change when dividing by 2?
5. How do we graph quadratic inequalities to visualize the solution?

Tip: Always check the direction of the inequality and whether it flips when you multiply or divide by a negative number!