Solve the inequality : 14x^2

We have to solve the inequality 14x^2 < 13x -
3

Now 14x^2 < 13x -
3

subtract 13x - 3 from both the
sides

=> 14x^2 - 13x + 3 <
0

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

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

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

Now for the product of the factors to be less than 0 only
one of them should be negative and the other
positive.

Therefore first
taking:

(7x - 3) <0  and ( 2x -1)
>0

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

But x cannot be less than 3/7 and greater than 1/2 at
the same time, so we can't find valid values here.

Let's
take (7x - 3) > 0  and ( 2x -1) <0

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

This gives valid values
in the range (3/7 , 1/2)

Therefore the set in
which the values of x lie is (3/7 , 1/2).