Solve the quadratic equation $6{{x}^{2}}-x-2=0$ for the value of x.

Hint: Here, we will first find the discriminant of the given quadratic equation by the formula $d={{b}^{2}}-4ac$ and we will check whether it is greater than or equal to zero. Then we will find the values of x using the quadratic formula.

Complete step-by-step answer:

We know that the general form of a quadratic equation is given as $a{{x}^{2}}+bx+c=0$, here x represents an unknown or variable and a is not equal to zero.
For solving a quadratic equation, we have several methods but, here we will the quadratic formula which is as follows:
For a general quadratic equation of the form $a{{x}^{2}}+bx+c=0$, according to the quadratic formula the value of x is given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, the $\pm $ sign indicates that the quadratic equation has two solutions.
The expression ${{b}^{2}}-4ac$ is called the discriminant of the quadratic equation. It tells us whether the quadratic equation has two solutions, one solution or no solution as per the following conditions:
(1) +ve value of discriminant tells us that the quadratic equation has two distinct real solutions.
(2) –ve value of discriminant tells us that the quadratic equation has no real solutions.
(3) If the value of discriminant is 0, it tells us that the quadratic equation has repeated real solutions, i.e. one solution.
Here, the equation given to us is:
$6{{x}^{2}}-x-2=0$
The discriminant of this quadratic equation is :
$d={{b}^{2}}-4ac={{\left( -1 \right)}^{2}}-4\times 6\times \left( -2 \right)=1+48=49$
Since, it is greater than 0, so the equation has two real and distinct solutions given as:
${{x}_{1}}=\dfrac{-\left( -1 \right)\pm \sqrt{49}}{2\times 6}=\dfrac{1+7}{12}=\dfrac{8}{12}=\dfrac{2}{3}$
And,
${{x}_{2}}=\dfrac{-\left( -1 \right)\pm \sqrt{49}}{2\times 6}=\dfrac{1-7}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}$
Hence, the solution of the given quadratic equation is $x=\dfrac{2}{3},\dfrac{-1}{2}$.

Note: Students should note here that we are getting two values of x because the degree of the equation is 2. Students must remember the formula for finding the discriminant and roots of a quadratic equation.