We are given the inequality:

$\lvert x + 1 \rvert^2 - 7 \lvert x + 1 \rvert + 10 < 0$

Step 1: Substitution

Let $u = \lvert x + 1 \rvert$. This simplifies the inequality to:

$u^2 - 7u + 10 < 0$

Step 2: Solve the quadratic inequality

Now, we need to solve the quadratic inequality:

$u^2 - 7u + 10 < 0$

First, solve the corresponding quadratic equation:

$u^2 - 7u + 10 = 0$

Using the quadratic formula:

$u = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)}$ $u = \frac{7 \pm \sqrt{49 - 40}}{2}$ $u = \frac{7 \pm \sqrt{9}}{2}$ $u = \frac{7 \pm 3}{2}$

Thus, the solutions are:

$u = \frac{7 + 3}{2} = 5 \quad \text{and} \quad u = \frac{7 - 3}{2} = 2$

So, the quadratic equation factors as:

$(u - 5)(u - 2) = 0$

Step 3: Analyze the inequality

The inequality becomes:

$(u - 5)(u - 2) < 0$

This product is negative when $u$ is between 2 and 5. Therefore:

$2 < u < 5$

Step 4: Revert substitution

Recall that $u = \lvert x + 1 \rvert$, so we have:

$2 < \lvert x + 1 \rvert < 5$

This inequality can be split into two cases:

1. $x + 1 > 2$, which simplifies to $x > 1$
2. $-(x + 1) > 2$, which simplifies to $x < -3$

Therefore, the solution for $x$ is:

$x < -3 \quad \text{or} \quad x > 1$

Final Answer:

The solution to the inequality is:

$x \in (-\infty, -3) \cup (1, \infty)$

Would you like more details or have any further questions?

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Here are five related questions to explore further:

1. How do you solve quadratic inequalities in general?
2. What are the properties of absolute value inequalities?
3. Can you solve an inequality with different quadratic forms?
4. How does the quadratic formula help in solving inequalities?
5. How do absolute value transformations affect inequalities?

Tip: Always remember to check for negative ranges when solving absolute value inequalities!