Question 597391
<pre>

I think the following is more in line with what your teacher has in mind:
     _____________
y = &#8730;2x² + 5x - 12

2x² + 5x - 12 &#8807; 0

(x + 4)(2x - 3) &#8807; 0

Now we find critical numbers, which are the 
zeros of the left side, by setting each factor
equal to 0:

x + 4 = 0;  2x - 3 = 0
    x = -4;      x = 3/2 = 1 1/2

We place these critical number on a number line: 

<========&#9899;--------------------&#9899;--------------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5 

We use darkened circles instead of open circles because the 
critical numbers are themselves solutions in this case since 
the inequality is &#8807; rather that >.

There are three intervals to test for solutions to the
inequality, the intervals between and beyond the critical
numbers

(-oo,-4), (-4, 3/2), and (3/2,oo).  Or if you haven't had
interval notation, then they are these inequalities:

x &#8806; -4,  -4 &#8806; x &#8806; 3/2,  x &#8807; 3/2

Choose a test point in the interval (-oo,-4), say -5.
Substitute 5 into the inequality:

    (x + 4)(2x - 3) &#8807; 0
(-5 + 4)(2(-5) - 3) &#8807; 0 
      (-1)(-10 - 3) &#8807; 0
          (-1)(-13) &#8807; 0
                 13 &#8807; 0

That is true, so (-oo,-4) is part of the solution, so we
shade that interval on the number line:

 Choose a test point in the interval (-4,3/2), say 0.
Substitute 0 into the inequality:

    (x + 4)(2x - 3) &#8807; 0
  (0 + 4)(2(0) - 3) &#8807; 0 
         (4)(0 - 3) &#8807; 0
            (4)(-3) &#8807; 0
                -12 &#8807; 0

That is false, so (-4,3/2) is NOT part of the solution, so we
DO NOT shade that interval on the number line, so we still have 
jus:t

<========&#9899;--------------------&#9899;--------------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5 

Choose a test point in the interval (3/2,-oo), say 2.
Substitute 2 into the inequality:

    (x + 4)(2x - 3) &#8807; 0
  (2 + 4)(2(2) - 3) &#8807; 0 
         (6)(4 - 3) &#8807; 0
             (6)(1) &#8807; 0
                  6 &#8807; 0

That is true, so (3/2,oo) is part of the solution, so we
shade that interval on the number line:

<========&#9899;--------------------&#9899;==============>
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5


So the domain written in interval notation is

(-oo,-4] U [3/2, oo)

or as an inequality the domain is written this way:

x &#8806; -4 OR x &#8807; 3/2

Edwin</pre>