To solve the inequality:

$x^2 + (x + 3)^2 - 1 - 2x > 0$

We can follow these steps:

1. Expand the square $(x + 3)^2$: $(x + 3)^2 = x^2 + 6x + 9$

2. Substitute this into the inequality: $x^2 + (x^2 + 6x + 9) - 1 - 2x > 0$

3. Simplify by combining like terms: $x^2 + x^2 + 6x + 9 - 1 - 2x > 0$ $2x^2 + 4x + 8 > 0$

4. Simplify the inequality further by dividing the entire inequality by 2: $x^2 + 2x + 4 > 0$

5. Analyze the quadratic expression $x^2 + 2x + 4 > 0$.

The discriminant $\Delta$ of the quadratic expression is calculated as follows: $\Delta = b^2 - 4ac = 2^2 - 4(1)(4) = 4 - 16 = -12$

Since the discriminant is negative ($\Delta < 0$), the quadratic has no real roots, and the parabola opens upwards (since the coefficient of $x^2$ is positive). Therefore, the expression $x^2 + 2x + 4$ is always positive for all real values of $x$.

6. Conclusion: $x^2 + 2x + 4 > 0$ holds for all $x \in \mathbb{R}$.

Thus, the solution to the inequality is: $\boxed{x \in \mathbb{R}}$

This means the inequality is true for all real numbers $x$.

Would you like more details on any step, or do you have any further questions?

Here are 5 related questions to consider:

1. How do you solve quadratic inequalities with real roots?
2. What is the significance of the discriminant in solving quadratic inequalities?
3. Can you give an example where the inequality has a different form, such as $x^2 - 2x + 1 > 0$?
4. How do you interpret the solution when the discriminant is positive?
5. How would the solution change if the parabola opened downwards?

Tip: When solving quadratic inequalities, always consider the sign of the discriminant to determine if the expression is always positive, always negative, or changes sign.