3x²-8x > 3
   3x²-8x-3 > 0
(x-3)(3x+1) > 0
Critical numbers are the zeroes of
the factors, 3, -1/3.

We graph the critical numbers on a number line.
Since the inequality symbols are > and not ,
the critical numbers cannot be solutions, so we draw 
open circles at each critical number.

-----------o--------------------o-----------
-2    -1  0     1     2     3     4     5

We test each interval.

We pick a test point on 
We pick -1 and substitute it in

   (x-3)(3x+1) > 0
  (-1-3)(3∙-1+1) > 0
    (-4)(-3+1) > 0
      (-4)(-2) > 0
             8 > 0

That's true so we shade that interval on
the number line:

<==========o--------------------o-----------
-2    -1  0     1     2     3     4     5  
Show that it has two different rational roots.

Next we pick a test point on 
We pick 0 and substitute it in

   (x-3)(3x+1) > 0
  (0-3)(3∙0+1) > 0
       (-3)(1) > 0
            -3 > 0

That's false so we DO NOT shade that interval on
the number line.  We still have this:

<==========o--------------------o-----------
-2    -1  0     1     2     3     4     5

We test each interval.

We pick a test point on 
We pick 4 and substitute it in

   (x-3)(3x+1) > 0
(4-3)(3∙4+1) > 0
     (1)(12+1) > 0
       (1)(13) > 0
            13 > 0

That's true so we shade that interval on
the number line:

<==========o--------------------o==========>
-2    -1  0     1     2     3     4     5

So the solution in interval notation is

  

----------------

[That's funny for we already had to solve it to
get the critical values in the first problem. :) ]

But pretending we hadn't, we would just show that
the discriminant is a perfect square:

Discriminant = b²-4ac = (-8)²-4(3)(-3) = 64+36 = 100
which is 10², a perfect square, so we know that the
quadratic has two rational roots. 

Edwin