 So let's try to solve the inequality x squared minus x minus 12 less than 0. So remember we solve inequalities by solving the corresponding equality x squared minus x minus 12 equals 0, and this is a quadratic equation that we can try to solve. We could try to factor, but remember factoring requires knowing every possible factor then using trial and error to find one that works. And remember factoring doesn't always work. Sometimes we can't factor. Fortunately, the quadratic formula always works, and what this means is we won't waste time trying to factor. So we'll go right to using the quadratic formula, which gives us our solutions 4 or negative 3. And if it's all written down, it didn't happen. Remember these solutions to the equality are what we call the critical values, and so we say that the critical values are x equals 4 or x equals negative 3. So the critical values partition the number line into three intervals. Since they're the values that make the equation true, but the inequality is strict, then we must exclude the critical values, and we indicate this by graphing these critical values using open circles. So we have three intervals, we need to test a point in each interval. In this interval on the left, we might let x equal negative 1 million, and x squared is going to be a large positive number, x squared minus x is going to be a large positive number, and x squared minus x minus 12 is going to be a large positive number, and so this inequality is false. And since it's false, that means we need to exclude this left interval. Next, we have this interval in the middle, and we see that 0 is a convenient value right inside the intervals. So let's use that as our test point. If x equals 0, then the question you've got to ask yourself is, is it true that x squared minus x minus 12 is less than 0? Equals means replaceable, so any place we see x will replace it with 0. We'll do a little arithmetic, and the question you've got to ask yourself is, is negative 12 less than 0? And this is true, so we should include this middle interval. Our last interval over on the right, we'll pick an x value like 10 lots, and we can substitute this value of x into our inequality, and we get a false statement, and so this right interval is not included in the solution set. And while the graph is a good way to represent the solution, we should also write this in interval notation. The solution set is just this middle portion, which goes from negative 3 up to 4, not including either end point. And so in interval notation, we write this as...