`-3<x^2+3x-1<3`

Now we need to break this up into 2 parts:

`-3<x^2+3x-1`

`0<x^2+3x+2`

`x^2+3x+2>0`

`(x+2)(x+1)>0`

Critical values for the left side are:

`x = -2 and x = -1`.

This gives:

`x < -2 and x > -1`.

`x^2+3x-1<3`

`x^2+3x-4<0`

`(x+4)(x-1)<0`

Critical values:

`x = -4` and `x = 1`

This gives:

`-4 < x < 1`

It's easier to graph these together on one axis to decide the final result:

These 3 regions intersect in the following 2 places:

This gives us our final solution: −4 < *x* < −2 and −1 < *x* < 1.

Remember to **check** numbers inside and outside the given solution region to make sure it works.

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