Question 62263
Solve {{{3x^2-x-4<=0}}}   Factor first.
{{{3x^2-4x+3x-4<=0}}}
{{{(3x^2-4x)+(3x-4)<=0}}}
{{{x(3x-4)+1(3x-4)<=0}}}
{{{(x+1)(3x-4)<=0}}}  Find the critical numbers that make this equal 0.
x+1=0 and 3x-4=0
x+1-1=0-1 and 3x-4+4=0+4
x+0=-1 and 3x-0=4
x=-1 and 3x=4
x=-1 and 3x/3=4/3
x=-1 and x=4/3  The critical numbers tell you the intervals to test.
(-infinity,-1], [-1,4/3],[4/3,infinity)
Choose a test point in the interval, if the statement is true, then that interval is part of the solution, if not, it's not.
For (-infinity,-1] test -2.
{{{(-2+1)(3(-2)-4)<=0}}}
{{{(-1)(-6-4)<=0}}}
{{{(-1)(-10)<=0}}}
{{{20<=0}}}  This is false, so this interval is not part of the solution.
:
For [-1,4/3] test 0.
{{{(0+1)(3(0)-4)<=0}}}
{{{(1)(0-4)<=0}}}
{{{1(-4)<=0}}}
{{{-4<=0}}}  This is true, so the interval [-1,4/3] is part of the solution.
:
For [4/3,infinity) test 2
{{{(2+1)(3(2)-4)<=0}}}
{{{(3)(6-4)<=0}}}
{{{3(2)<=0}}}
{{{6<=0}}}  This is false, so this interval is not part of the solution.
:
Therefore, the solution set is: [-1,4/3].
:
Note that if the inquality symbol was reversed the solution set would be the two intervals that we rejected.  It is possible to have more than one interval as your solution set.
Happy Calculating!!!