Distribute the negative through the terms of the binomial.

Subtract  on both sides.

Add 18 on both sides.

Divide by 13 on both sides.

The answer is:  

The first thing we can do is distribute the  and the  into their respective terms:

Now we can start to simplify by gathering like terms.  Remember, if we multiply or divide by a negative number, we change the direction of the inequality:

It can be tricky because there's a negative sign in the equation, but we never end up multiplying or dividing by a negative, so there's no need to change the direction of the inequality.  We simply divide by  and multiply by :

Remember, anything we do to one side of the inequality, we must also do to the other two sides.  We can start by adding one to all three sides:

And now we divide each side by two:

The first thing we can do is distribute the negative sign in the middle term.  Because we're not multiplying or dividing the entire inequality by a negative, we don't have the change the direction of the inequality signs:

Now we can subtract  from the inequality:

And finally, we can multiply by .  Note, this time we're multiplying the entire inequality by a negative, so we have to change the direction of the inequality signs:

We start by distributing the :

Now we can add , and subtract an  from each side of the equation:

We didn't multiply or divide by a negative number, so we don't have to reverse the direction of the inequality sign.

Start by finding the roots of the equation by changing the inequality to an equal sign.

Now, make a number line with the two roots:

Pick a number less than  and plug it into the inequality to see if it holds.

For ,

 is clearly not true. The solution set cannot be .

Next, pick a number between .

For ,

 is true so the solution set must include .

Finally, pick a number greater than .

For ,

 is clearly not the so the solution set cannot be .

Thus, the solution set for this inequality is .

First, find the zeroes of the numerator and the denominator. This will give the boundary points of the intervals to be tested.

; 

Since the numerator may be equal to 0,  and  are included as solutions. However, since the denominator may not be equal to 0,  is excluded as a solution. 

Now, test each of four intervals for inclusion in the solution set by substituting one test value from each:

Let's test :

This is false, so  is excluded from the solution set.

Let's test :

This is true, so  is included in the solution set.

Let's test :

This is false, so  is excluded from the solution set.

Let's test :

This is true, so  is included in the solution set.

The solution set is therefore .

Add 6 on both sides.

Divide by five on both sides.

The answer is:  

Add six on both sides.

Divide by three on both sides.

The answer is: