The inequality is .
can be written as
.
Let .
.
Therefore .
is in the form of
.
The factored form of is
.
Therefore has
real zero at
.
Create a sign chart using the value .
Note : The solid circle denote that the value are included in the solution set.
\Observe the sign chart :
\The set of value of denoted in pink color represents the solution set.
Determine whether is positive or negative on the test intervals.
Test intervals are and
.
\
If ,then
.
If , then
.
The solutions of are
values such that
is positive or equal to
.
From the sign chart, you can see that the solution set is .
The solution set is .
The solution set of is
.