Which are the elements of the set A if they are solution of the inequality x^2 - 13x/3 - 10/3 > 0 ?

To determine the elements of the set, we'll sove the
inequality.

For this purpose, will solve first the
equation:

x^2 - 13x/3 - 10/3 =
0

3x^2 - 13x - 10 = 0

After
finding the roots of the equation, we could write the expression in a factored form
as:

3(x-x1)(x-x2)>0

So,
let's apply the quadratic formula to calculate the
roots:

x1 =
[13+sqrt(169-120)]/6

x1 =
(13+sqrt49)/6

x1 =
(13+7)/6

x1 =
10/3

x2 =
(13-7)/6

x2 =
6/6

x2 =
1

The inequality will be written
as:

3(x - 10/3)(x - 1) >
0

We'll divide by 3, both
sides:

(x - 10/3)(x - 1) >
0

Now, we'll discuss the
inequality:

- the product is positive if the factors are
both positive:

x -
10/3>0

We'll add 10/3 both
sides:

x >
10/3

and

x - 1 >
0

We'll add 1 both sides:

x
> 1

So, x belongs to the interval (10/3 ,
+inf.)

- the product is positive if the factors are both
negative:

x - 10/3 <
0

We'll add 10/3 both sides:

x
< 10/3

x - 1 <
0

We'll add 1 both sides:

x
< 1

So, x belongs to the interval (
-inf.,1)

Finally, the solution set of the inequality is the
union of the sets  above:

( -inf.,1) U (10/3
, +inf.)

So, the set A is the
union of intervals:

A =
{( -inf.,1) U (10/3 ,
+inf.)}