How do you solve the inequality: `|x^2-9x+15|>1`?

By the definition of the absolute value of a number, this is equivalent to stating that

`x^2-9x+15 > 1`
or
`x^2-9x+15 < -1`

We will solve each inequality separately.

For the first inequality, subtracting `1` from each side, we obtain

`x^2-9x+14 > 0`

`=> (x-2)(x-7) > 0`

`=> x-2 > 0` and `x-7 > 0`
or
`x-2 < 0` and `x-7 < 0`

`=>x>7`
or
`x<2`

`=> x in (-oo,2)uu(7,oo)`

For the second inequality, we can add `1` to both sides and use the quadratic formula to factor the left hand side.

`x^2-9x+16 < 0`

`=> (x-(9+sqrt(17))/2)(x-(9-sqrt(17))/2) < 0`

Because `(x-(9+sqrt(17))/2) < (x-(9-sqrt(17))/2)` and the inequality holds only when one of factors is negative and the other is positive, we must have that

`(x-(9+sqrt(17))/2) < 0` and `(x-(9-sqrt(17))/2) > 0`

`=> (9-sqrt(17))/2 < x < (9+sqrt(17))/2`

`=> x in ((9-sqrt(17))/2, (9+sqrt(17))/2)`

To get our final solution set, we can take the union of both solutions we found above to obtain

`x in (-oo,2) uu ((9-sqrt(17))/2, (9+sqrt(17))/2)uu(7,oo)`

Note that this matches what we would expect from the graph, as `|x^2-9x+15|-1` is negative on only two small intervals:

graph{|x^2-9x+15|-1 [-5.205, 14.795, -3.24, 6.76]}