x^2-2x-8<0

To solve the inequality \( x^2 - 2x - 8 < 0 \), we first need to find the roots of the corresponding equation \( x^2 - 2x - 8 = 0 \).

We can use the quadratic formula:

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

Here, \( a = 1 \), \( b = -2 \), and \( c = -8 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36 \]

Now, substituting into the quadratic formula:

\[ x = \frac{-(-2) \pm \sqrt{36}}{2(1)} = \frac{2 \pm 6}{2} \]

This gives us two solutions:

1. \( x = \frac{2 + 6}{2} = \frac{8}{2} = 4 \)
2. \( x = \frac{2 - 6}{2} = \frac{-4}{2} = -2 \)

So, the roots of the equation are \( x = -2 \) and \( x = 4 \).

Next, we need to determine the intervals where \( x^2 - 2x - 8 < 0 \). The roots \( x = -2 \) and \( x = 4 \) divide the number line into three intervals:

1. \( (-\infty, -2) \)
2. \( (-2, 4) \)
3. \( (4, \infty) \)

We can test a point from each interval to see where the inequality holds:

1. For the interval \( (-\infty, -2) \), pick \( x = -3 \): \[ (-3)^2 - 2(-3) - 8 = 9 + 6 - 8 = 7 \quad (\text{not } < 0) \]

2. For the interval \( (-2, 4) \), pick \( x = 0 \): \[ 0^2 - 2(0) - 8 = -8 \quad (< 0) \]

3. For the interval \( (4, \infty) \), pick \( x = 5 \): \[ 5^2 - 2(5) - 8 = 25 - 10 - 8 = 7 \quad (\text{not } < 0) \]

The inequality \( x^2 - 2x - 8 < 0 \) holds for the interval:

\[ (-2, 4) \]

Thus, the solution to the inequality \( x^2 - 2x - 8 < 0 \) is:

\[ \boxed{(-2, 4)} \]