How do you solve using the completing the square method \[3{{x}^{2}}-7x-6=0\]?

Hint: In this problem, we have to find the value of x by solving the given quadratic equation in completing the square method. For this, we can first convert the given quadratic equation into a complete square equation by algebraic whole square formula. we can add the missing term on both sides to get a complete square equation on the left hand side, then we will get a whole square form, for which we can take square root on both sides to get the value of x.

Complete step by step solution:
We know that the given quadratic equation to be solved is,
\[3{{x}^{2}}-7x-6=0\] ……. (1)
Now we can add 3 on both sides in the above equation (1), we get
\[\Rightarrow 3{{x}^{2}}-7x-6+6=6\]
Now we can divide on both sides to get a perfect square equation, we get
\[\begin{align}
  & \Rightarrow \dfrac{3{{x}^{2}}}{3}-\dfrac{7x}{3}=\dfrac{6}{3} \\
 & \Rightarrow {{x}^{2}}-\dfrac{7}{3}x=2.......(2) \\
\end{align}\]
We can now take the first two terms \[{{x}^{2}}-\dfrac{7}{3}x\]
We know that,
\[{{\left( x-a \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}\]
We can see that,
\[\begin{align}
  & \Rightarrow -2a=-\dfrac{7}{3} \\
 & \Rightarrow a=\dfrac{7}{6} \\
 & \Rightarrow {{a}^{2}}={{\left( \dfrac{7}{6} \right)}^{2}}=\dfrac{49}{36} \\
\end{align}\]
We can add the above value in both the left-hand side and the right-hand side of equation (2), to get a perfect square equation, we get
\[\Rightarrow {{x}^{2}}-\dfrac{7}{3}x+\dfrac{49}{36}=2+\dfrac{49}{36}\]
We can write the left-hand side in whole square form as it is a perfect square equation, we get
\[\Rightarrow {{\left( x-\dfrac{7}{6} \right)}^{2}}=\dfrac{2\times 36+49}{36}=\dfrac{121}{36}={{\left( \dfrac{11}{6} \right)}^{2}}\]
Now we can take square root on both sides, we get
\[\begin{align}
  & \Rightarrow x-\dfrac{7}{6}=\pm \dfrac{11}{6} \\
 & \Rightarrow x=\dfrac{7}{6}\pm \dfrac{11}{6} \\
\end{align}\]
Now we can simplify the above step, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{7+11}{6}=\dfrac{18}{6}=3 \\
 & \Rightarrow x=\dfrac{7-11}{6}=\dfrac{-2}{3} \\
\end{align}\]
Therefore, the value of \[x=3,-\dfrac{2}{3}\].

Note: We can now verify to check whether the values we got are correct or not. We can substitute the values of x in the equation to check.
We can now take the equation (1) and substitute the value of \[x=3,-\dfrac{2}{3}\], we get
When x = 3,
\[\begin{align}
  & \Rightarrow 3{{\left( 3 \right)}^{2}}-7\left( 3 \right)-6=0 \\
 & \Rightarrow 27-27=0 \\
\end{align}\]
When \[x=\dfrac{-2}{3}\],
\[\begin{align}
  & \Rightarrow 3{{\left( \dfrac{-2}{3} \right)}^{2}}-7\left( \dfrac{-2}{3} \right)-6=0 \\
 & \Rightarrow \dfrac{4}{3}-\dfrac{4}{3}=0 \\
\end{align}\]
Therefore, we can verify that the value for is \[x=3,-\dfrac{2}{3}\].