We get 0 on the right:



After getting LCD and simplifying, we have:



To find all critical values, we set the numerator and denominator
equal to 0 and solve for x:

We set the numerator = 0 and solve for x:

, which we solve by the quadratic formula 
and get 

 which are critical numbers 
approximately -11.8 and -6.2.

We set the denominator = 0,  and solve for 
x and get -1,-8, and -9.

So we place all the critical numbers on a number line, guessing
about where -11.8 and -6.2 should go between integers.

-----------o----------o---o------o--------------------o--------
-14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1

We pick the most convenient test value in each interval, and
substitute it into 
and if we get a non-positive number, we include that interval
in the solution.  If we get a positive number we do not.

For interval 
we choose test value -12, substitute it, and get a negative
number, so we include that interval.

For interval 
we choose test value -10, substitute it, and get a positive
number, so we do not include that interval.

For interval 
we choose test value -8.5, substitute it, and get a negative
number, so we include that interval.

For interval 
we choose test value -7, substitute it, and get a positive
number, so we do not include that interval.

For interval 
we choose test value -2, substitute it, and get a negative
number, so we include that interval.

For interval 
we choose test value 0, substitute it, and get a positive
number, so we do not include that interval.

Next, we include the critical values which cause the numerator
to be 0, which are .

Thus the solution is 



Edwin