So the signs change when . We consider, in turn, the four ranges . I'll do the first two, and leave you to look at the last two.
I'll leave you to check that this can be reduced to
Completing the squareNow for . So we want values of for which
So there's part of the solution set. (I said it was awkward!)
Now we look at the next range, .
In a similar way, we get this time:which can be reduced to
This time the denominator is negative, so we want values of for whichand clearly there are none. So there are no values of in the range .
I'll leave you to look at the last two ranges of in the same way. I reckon the final answer is .