To find the roots set the function to zero: 

 ,

                                                      (1)

Apply the quadratic formula:

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Reminder

Recall that for a quadratic   the general formula for the solution in terms of the constant coefficients is given by:  

                                                         (2) 

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Use equation (2) to write a solution for equation (1). 

If we simplify the right-hand term in the numerator we obtain:

So now we have for , 

After all the cancellations in the expression above we obtain:  

Therefore, the solution set for this equation is:  

The roots are the values of  for which: 

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Reminder

Recall that for a quadratic   the general formula for the solution in terms of the constant coefficients is given by:  

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Use the quadratic formula to find the roots. 

Notice that  is not a real number, and therefore the roots will be complex numbers. 

Using the definition of the imaginary unit  we can rewrite  as follows, 

Now we can write the solutions to this problem in the form: 

For this problem

a=1, coefficient of x^2 term

b=9, the coefficient of the x term

c=15, the constant term

solving the expression shows the roots of -6.79 and -2.21

Recall that the quadratic formula is defined as:

For this question, the variables are as follows:

Substituting these values into the equation, you get:

Use a calculator to determine the final values.

The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal.

Negative x^2's are hard to work with, so we multiply through by -1.

Now we can factor easily.

By the zero product property, each of these factors will be equal to 0.

Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality.

Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as:

First we want to rewrite the quadratic in standard form:

Now we want to set it = 0 and factor and solve like normal.

Using the zero product property, both factors produce a zero:

So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.

Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.

Ergo, .

The discriminant of a quadratic equation in  form is equal to . The given equation is not in that form however, so we must first multiply it out to get it into that form. We therefore obtain:

We therefore have , , and . Our discriminant is therefore:

The correct answer is therefore 

1. Rewrite the equation in standard form.

2. Set the equation equal to  and solve by factoring.

So,  and  are our zeroes.

3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing  by plugging it into your original inequality.)

Because the above statement is true, the solution is the interval between  and .

Combine like terms first.

Factor

The zeroes are 3 and 8 so a number line can be divided into 3 sections.

X<3 works, 3<x<8 does not work, and x>8 works

Start by setting the inequality to zero and by solving for .

Now, plot these two points on to a number line.

Notice that these two numbers effectively divide up the number line into three regions:

, , and 

Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.

For , let 

Since this  is not less than , the solution to this inequality cannot lie in this region.

For , let .

Since this will make the inequality true, the solution can lie in this region.

Finally, for , let 

Since this number is not less than zero, the solution cannot lie in this region.

Thus, the solution to this inequality is