How do you solve the inequality $ 6{x^2} - 5x > 6 $ ?

Hint: We have to solve an inequation. The given inequation is a quadratic inequation in one variable. The solution for an inequation is a range of values $ x $ and not just two values of $ x $ which satisfies the given inequation, i.e. such a range of values for which $ LHS > RHS $ .

Complete step by step solution:
We have to solve the inequality $ 6{x^2} - 5x > 6 $ .
Solving the given inequality means finding the range of values of $ x $ for which $ LHS > RHS $ .
First we transfer all the terms to the left side of the inequation. For this we subtract $ 6 $ from both sides.
 $
   \Rightarrow 6{x^2} - 5x - 6 > 6 - 6 \\
   \Rightarrow 6{x^2} - 5x - 6 > 0 \;
  $
Now we can factorize the quadratic expression in the left side. This we can do by splitting the middle term.
 $
  6{x^2} - 5x - 6 > 0 \\
   \Rightarrow 6{x^2} - 9x + 4x - 6 > 0 \\
   \Rightarrow 3x\left( {2x - 3} \right) + 2\left( {2x - 3} \right) > 0 \\
   \Rightarrow \left( {3x + 2} \right)\left( {2x - 3} \right) > 0 \;
  $
Thus, we get $ \left( {3x + 2} \right)\left( {2x - 3} \right) > 0 $ . The product of both the factors is positive. This means that either both the factors can be positive or both the factors can be negative.
Case 1:
 $
  \left( {3x + 2} \right) > 0\;\;\;and\;\;\;\left( {2x - 3} \right) > 0 \\
   \Rightarrow 3x > - 2\;\;\;and\;\;\;2x > 3 \\
   \Rightarrow x > - \dfrac{2}{3}\;\;\;and\;\;\;x > \dfrac{3}{2} \;
  $
For case 1 we get the common range $ x > \dfrac{3}{2} $ .
Case 2:
 $
  \left( {3x + 2} \right) < 0\;\;\;and\;\;\;\left( {2x - 3} \right) < 0 \\
   \Rightarrow 3x < - 2\;\;\;and\;\;\;2x < 3 \\
   \Rightarrow x < - \dfrac{2}{3}\;\;\;and\;\;\;x < \dfrac{3}{2} \;
  $
For case 2 we get the common range $ x < - \dfrac{2}{3} $ .
The final range of $ x $ for solution would be the combined range of case 1 and case 2 as both the cases are independent.
Thus, the final solution is $ x < - \dfrac{2}{3} $ or $ x > \dfrac{3}{2} $ . This we can also write as $ x \in \left( { - \infty , - \dfrac{2}{3}} \right) \cup \left( {\dfrac{3}{2},\infty } \right) $ .
Hence, the solution for the given inequality is $ x \in \left( { - \infty , - \dfrac{2}{3}} \right) \cup \left( {\dfrac{3}{2},\infty } \right) $ .
So, the correct answer is “ $ x \in \left( { - \infty , - \dfrac{2}{3}} \right) \cup \left( {\dfrac{3}{2},\infty } \right) $ ”.

Note: Since we were given to solve an inequality we got a range of values as a solution and not just two values of $ x $ . While solving the inequation we divided the condition into two cases which are independent to each other. We can check our solution by putting any value from the range in the given inequation and check whether the inequality holds true.