How do you solve `c^ { 2} - c - 20< 0`?

Okay, the first step is to factor this inequality:

`c^2-c-20<0`

`(c-5)(c+4)<0`

Next, let's use a sign chart. We can draw two dashed, vertical lines at `x=5` and `x=-4`. You can see that we have split the graph into three sections: negative infinity to -4; -4 to 5; and 5 to infinity. Now, we can choose any number less that -4 (let's say -10). If we plug it in, we get:

`(-10-5)(-10+4)=(-15)(-6)=90`

We can see that any number from negative infinity to -4 is positive if we plug it in the function. Now, let's choose a number between -4 and 5 (let's use 0):

`(0-5)(0+4)=(-5)(4)=-20`

We can see that any number from -4 to 5 is negative if we plug it in the function. Finally, let's choose a number from 5 to infinity (let's use 10):

`(10-5)(10+4)=(5)(14)=70`

We can see that any number from 5 to infinity is positive if we plug it into the function. Since the function calls for numbers less than 0, we can clearly see that the answer is:

`-4 < c < 5`

We don't include `-4` and `5` because those numbers make the function equal to zero, and we only want values less than 0.

Another (easier) way to solve this function is to graph it:

graph{x^2-x-20 [-13.56, 14.92, -7.01, 7.23]}

The two x-intercepts are `-4` and `5`. The graph goes under the x-axis between the two x-intercepts, meaning the function is negative between the two x-intercepts (which is what we got originally). Hope this answer helps!