14x^2 - 13x + 3 > 0

First let us factor:

==> (7x -3)(2x-1) > 0

Then we have two cases.

Case 1:

(7x-3) > 0 and (2x-1)>0

==> x > 3/7 AND x > 1/2

==> x > 1/2

Case2:

7x-3 < 0 and 2x-1 < 0

==> x < 3/7 AND x < 1/2

==> x < 3/7

Then. the solutin is:

**x belongs to (-inf, 3/7) U (1/2, inf)**

**OR:**

**x = R - [3/7, 1/2]**

The inequation 14x^2 - 13x+3 >0 can be written as:

14x^2 - 13x+3 >0

=> 14x^2 - 7x - 6x +3 >0

=> 7x ( 2x -1) -3 (2x -1) >0

=> (7x-3)(2x-1) >0

This is true if both (7x-3) and (2x-1) are positive

=> (7x-3)>0 and (2x-1) >0

=> x > 3/7 adn x > 1/2

=> x>1/2

Or if both (7x-3) and (2x-1) are negative

=> 0 >(7x-3) and 0 >(2x-1)

=> 3/7 > x and 1/2 > x

=> 3/7 >x

**Therefore we have the inequation is true for all values of x where x > 1/2 and x < 3/7**

To solve the inequation 14x^2-13x+3 > 0.

We first solve the equation 14x^2-13x+3 = 0.

14x^2-7x -6x+3 = 0

7x(2x-1)-3(2x-3) = 0

(2x-1)(7x-3) = 0

2x-1 = 0. Or 7x-3 = 0

x = 1/2 or x = 3/7.

3/7 < 1/2.

Therefore (7x-3)(2x-1) > when both factors are negative oe both factors are positive. This is possible whe x <3/7 or when x>1/2

Therefore 14x^2-13x+3 = (7x-3)(2x-1) > 0 when x < 3/7 or when x> 1/2.

We'll apply another method of solving the inequality.

We'll find the roots of the equation and we'll establish the rule: between the roots, the expression has the opposite sign to the sign of the coefficient of x^2. The expression will have the same sign with the sign of the coefficient of x^2, outside the roots.

14x^2 - 13x+3 = 0

Since it is a quadratic, we'll apply the quadratic formula.

x1 = [13+sqrt(169-168)]/2*14

x1 = 14/28

x1 = 1/2

x2 = 12/48

x2 = 1/4

**Since the coefficient of x^2 is positive, the expression will be positive outside the roots.**

**14x^2 - 13x+3 >0 for x belongs to the intervals:**

**(-infinite, 1/4) U (1/2, +infinite) **