(х+12)(х-6)<0

To solve the inequality \((x+12)(x-6)<0\), we first need to find the values of \(x\) where the product \((x+12)(x-6)\) is equal to 0. This occurs at the points where each factor equals zero:

1. \(x + 12 = 0 \implies x = -12\)
2. \(x - 6 = 0 \implies x = 6\)

Now, the critical points are \(x = -12\) and \(x = 6\). These points divide the number line into three intervals:

1. \( (-\infty, -12) \)
2. \( (-12, 6) \)
3. \( (6, \infty) \)

Next, we will test a point from each interval to determine where the product \((x+12)(x-6)\) is negative.

1. For the interval \((- \infty, -12)\), let's choose \(x = -13\): \[ (-13 + 12)(-13 - 6) = (-1)(-19) = 19 \quad (\text{positive}) \]

2. For the interval \((-12, 6)\), let's choose \(x = 0\): \[ (0 + 12)(0 - 6) = (12)(-6) = -72 \quad (\text{negative}) \]

3. For the interval \((6, \infty)\), let's choose \(x = 7\): \[ (7 + 12)(7 - 6) = (19)(1) = 19 \quad (\text{positive}) \]

Now we summarize the results:

• \((- \infty, -12)\): Positive
• \((-12, 6)\): Negative
• \((6, \infty)\): Positive

The inequality \((x+12)(x-6)<0\) holds only in the interval \((-12, 6)\).

Thus, the solution to the inequality \((x+12)(x-6)<0\) is:

\[ \boxed{(-12, 6)} \]