x2-5x-36<0

To solve the inequality \( x^2 - 5x - 36 < 0 \), we first need to find the roots of the corresponding equation \( x^2 - 5x - 36 = 0 \).

We can do this using the quadratic formula:

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

In our case, \( a = 1 \), \( b = -5 \), and \( c = -36 \). Plugging in these values, we have:

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (-5)^2 - 4 \cdot 1 \cdot (-36) = 25 + 144 = 169 \]

Now we can find \( x \):

\[ x = \frac{5 \pm \sqrt{169}}{2} \] \[ x = \frac{5 \pm 13}{2} \]

This gives us two solutions:

\[ x_1 = \frac{18}{2} = 9 \] \[ x_2 = \frac{-8}{2} = -4 \]

Now we have the roots \( x = 9 \) and \( x = -4 \).

Next, we need to determine the intervals on which the expression \( x^2 - 5x - 36 \) is less than 0. The critical points divide the real number line into three intervals:

1. \( (-\infty, -4) \)
2. \( (-4, 9) \)
3. \( (9, \infty) \)

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

1. For \( x = -5 \) (in the interval \( (-\infty, -4) \)): \[ (-5)^2 - 5(-5) - 36 = 25 + 25 - 36 = 14 > 0 \] (not part of the solution)

2. For \( x = 0 \) (in the interval \( (-4, 9) \)): \[ 0^2 - 5(0) - 36 = -36 < 0 \] (this interval is part of the solution)

3. For \( x = 10 \) (in the interval \( (9, \infty) \)): \[ 10^2 - 5(10) - 36 = 100 - 50 - 36 = 14 > 0 \] (not part of the solution)

From this testing, we find that the inequality \( x^2 - 5x - 36 < 0 \) holds in the interval \( (-4, 9) \).

Therefore, the solution to the inequality is:

\[ \boxed{(-4, 9)} \]