Solve $3{x^2} - \sqrt 6 x + 2 = 0$ .

Hint: Before you start solving the equation in the form $a{x^2} + bx + c = 0$ we need to determine the nature of the discriminant i.e, ${b^2} - 4ac$
If ${b^2} - 4ac < 0$ , then there will be no real solutions.
If ${b^2} - 4ac > 0$ , then there will be two real solutions.
If ${b^2} - 4ac = 0$ , then there will be one real solution.
If ${b^2} - 4ac \geqslant 0$ then proceed further to solve for x using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Stepwise Solution:
 Given: $3{x^2} - \sqrt 6 x + 2 = 0$.
Comparing given eqaution with the general form of quadratic equation $a{x^2} + bx + c = 0$ we get,
From the above equation a is $3$ , b is $ - \sqrt 6 $ and c is $2$ .
Discriminant ${b^2} - 4ac$ is as follows:
$ = {\left( { - \sqrt 6 } \right)^2} - 4 \times 3 \times 2$
$
   = 6 - 24 \\
   = - 18 \\
   \Rightarrow - 18 < 0 \\
$
As ${b^2} - 4ac < 0$, there will be no real solutions.

Note: In such types of questions which involve the concept of finding a solution to a given equation we will need to have knowledge about discriminant and formula to find the value of x. As calculations play a critical role in these kinds of questions we will need to be vigilant about that while solving.