3x^2 - 13x - 10 > 0

First let us factor:

( 3x^2 - 13 x -10 = (3x +2)(x-5)

==> (3x+2)(x-5) > 0

That means:

(3x+2) > 0 and (x-5)> 0

==> x > -2/3 and x > 5

==> x belongs to (5, inf).......(1)

ALso,

(3x+2) < 0 and (x-5)<0

==> x < -2/3 and x < 5

==> x belongs to (-inf, -2/3)......(2)

From (1) and (2) we have:

x belongs to (-inf, -2/3) U (5, inf)

f(x) = 3x^2 - 13x - 10 > 0

First, I solve the equation by the Diagonal Sum Method. Roots have opposite signs. There are 3 probable root pairs:

(-1/3, 10/1),(-2/1, 5/3),(-2/3, 5/1).

The third diagonal sum is (-2 + 15 = 13 = -b).

The 2 real roots are -2/3 and 5.

You may solve the quadratic inequalities by the Test Point Method. I solve it by the algebraic method because I don't have to draw the number line.

Beyween the 2 real roots -2/3 and 5, f(x) is negative (-) as opposite in sign to a = 3. So, f(x) is positive outside the interval (-2/3, 5). The answer or solution set is: (-inf, -2/3) U (5, +inf).

Check: The parabola f(x) is open upward. Between the 2 real roots, one part the parabola f(x) is below the x-axis. Outside the interval (-2/3, 5) the parabola graph is above the x-axis, meaning f(x) > 0.

3x^2-13x-10 = 0.To solve for x.

Solution:

3x^2-13x - 10 = 0

3x^2-13x = 10

3(x^2-13x/3) = 10. Divide by 3.

x^2-13x/3 = 10/3. Add (13/(3*2))^2 to both sides so that left becomes a perfect square.

x^2 -2*13x/(2*3) +(13/6)^2 = 10/3 +(13/6)^2 =

(x-13/6)^2 = (120+169)/6^2

x-13/6 = +or -sqrt(289/36)

x1 = 13/6+17/6 = 5

X2 =13/6-17/6 = -2/3.

To solve the inequality above, first we have to calculate the roots of the equation 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:

(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

x>10/3

and

x-1>0

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

x<10/3

x-1<0

x<1

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

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

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