Find the range of values of $x$ which satisfies the inequality. Find the range of values of $x$ which satisfies the inequality $(2x+1)(3x-1)<14$.
I have done more similar sums and I know how to solve it. I tried this one too but my answer doesn't matches the book's answer. 
I did in this way: Solving, i.e, After rearranging the equation and factorizing it, I get
$(3x+5)(2x-3)<0$
Is this right?
Anyways, then finding the range of values, I get: $x<-\frac{5}{3}$ or $ x>\frac{3}{2}$
But my book says the answer should be $-\frac{5}{3}<x<\frac{3}{2}$.
Did I do any mistake?
 A: Consider the inequality
$$ab\lt 0\tag 1$$
This inequality holds exactly when $a\gt 0, b\lt 0$ or when $a\lt 0, b\gt 0$, but not when $a\gt 0, b\gt 0$ or $a\lt 0,b\lt 0$.
In this question, the inequality has been reduced to
$$(3x+5)(2x-3)\lt0$$
Next it is necessary to find out the comparisons of $3x+5\lt 0$ and $2x-3\lt 0$, which have been found as $x\lt -\frac 53$ and $x\lt \frac 32$.
Further, we have that $-\frac 53\lt \frac 32$.  So what remains is to put this information together with the conditions satisfying $(1)$.  In particular, we have $2x-3\lt 3x+5$ when $x\gt -\frac 53$.  If $x\lt -\frac 53$ then both $2x-3$ and $3x+5$ are negative, which breaks the conditions on $(1)$, so we cannot have $x\lt -\frac 53$, and similarly if $x\gt \frac 32$ then both $2x-3$ and $3x+5$ are positive, which also breaks our conditions.  But if $-\frac 53\lt x\lt \frac 32$ then $2x-3$ is negative and $3x+5$ is positive, and the necessary conditions are met to satisfy our inequality.
A: Thinking purely geometrically, you might realize that your left hand side defines a parabola $y=(2x+1)(3x-1)$ opening up, while the right hand side is a constant.  Again, from a purely geometric point of view, this must be an open interval (possibly empty).

A: The inequality you obtained, $$(3x+5)(2x-3) < 0,$$ is correct.  The next step is to find where the LHS equals zero:  this gives you the $boundary$ of the intervals for which the inequality is satisfied.  You will find $x = -5/3$ and $x = 3/2$.  This means we should consider three intervals:  $x < -5/3$, $-5/3 < x < 3/2$, and $x > 3/2$.
Pick test points from each interval.  For example, pick $x = -2$.  Plug it into the inequality.  Is it true?  $(3(-2) + 5)(2(-2) - 3) = (-1)(-7) = 7 > 0$, so $x = -2$ does not satisfy the inequality.  Next, pick $x = 0$:  we get $(3(0)+5)(2(0)-3) = (5)(-3) = -15 < 0$, which does satisfy the inequality, so we know that $-5/3 < x < 3/2$ is a solution to the given inequality.  Finally, just to check, try picking $x = 2$.  Does it work?
A: You're basically looking for values when one of 3x+5,2x-3 is negative and the other one positive(or smaller then zero)
Since when $$x>\frac{3}{2} \implies (3x+5)(2x−3)>0\\x<-\frac{5}{3} \implies (3x+5)(2x−3)>0$$
In first case both 3x+5 and 2x-3 are positive,and in second both are negative
