To solve the inequality $2x^2 + 6x - 6 < x^2 + 2x$, we start by rearranging it into a standard form:

1. Subtract $x^2$ and $2x$ from both sides:

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

   This simplifies to:

   $x^2 + 4x - 6 < 0$

2. Next, we find the roots of the quadratic equation $x^2 + 4x - 6 = 0$ using the quadratic formula:

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

   Here, $a = 1$, $b = 4$, and $c = -6$:

   $x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}$

   $x = \frac{-4 \pm \sqrt{16 + 24}}{2}$

   $x = \frac{-4 \pm \sqrt{40}}{2}$

   $x = \frac{-4 \pm 2\sqrt{10}}{2}$

   $x = -2 \pm \sqrt{10}$

   Thus, the roots are $x_1 = -2 - \sqrt{10}$ and $x_2 = -2 + \sqrt{10}$.

3. The critical points divide the number line into intervals. We need to test each interval to see where the inequality $x^2 + 4x - 6 < 0$ holds.

   The intervals are:

   • $(-\infty, -2 - \sqrt{10})$
   • $(-2 - \sqrt{10}, -2 + \sqrt{10})$
   • $(-2 + \sqrt{10}, \infty)$

4. Choose test points in each interval:

   • For $x = -5$ in $(-\infty, -2 - \sqrt{10})$: $(-5)^2 + 4(-5) - 6 = 25 - 20 - 6 = -1 < 0 \quad \text{(true)}$

   • For $x = 0$ in $(-2 - \sqrt{10}, -2 + \sqrt{10})$: $0^2 + 4(0) - 6 = -6 < 0 \quad \text{(true)}$

   • For $x = 3$ in $(-2 + \sqrt{10}, \infty)$: $3^2 + 4(3) - 6 = 9 + 12 - 6 = 15 > 0 \quad \text{(false)}$

5. Therefore, the solution to the inequality $2x^2 + 6x - 6 < x^2 + 2x$ is:

   $x \in (-2 - \sqrt{10}, -2 + \sqrt{10})$

Summary

The solution is:

$(-2 - \sqrt{10}, -2 + \sqrt{10})$

Do you want details or have any questions? Here are some related questions:

1. What is the graphical representation of the inequality?
2. How do the roots affect the shape of the quadratic?
3. Can you explain the steps for solving a different type of inequality?
4. What is the significance of the discriminant in quadratic equations?
5. How do we determine the intervals for testing in inequalities?

Tip: Always check test points within each interval to confirm where the inequality holds true.