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.

solve each inequality 2x^2 + 6x - 6 < x^2 + 2x

Discover how to solve the inequality 2x^2 + 6x - 6 < x^2 + 2x by simplifying the expression and using the quadratic formula. This step-by-step solution includes finding critical points, interval testing, and understanding the behavior of quadratic inequalities. Ideal for Grades 9-11.

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