How do you solve $3{x^2} + 6x - 2 = 0$ by completing the square ?

Hint: Here we are given a quadratic equation. We will be applying completing the square method to find the variable x. Firstly, we have to make the coefficient of ${x^2}$ equal to 1. So, divide the whole equation by 3 and simplify. Then take the constant term to R.H.S. Add ${\left( {\dfrac{2}{2}} \right)^2}$ to both the sides of the quadratic equation and rearrange to make it complete square. Then we try to solve it to get in a required square form.

Complete step by step solution:
Given the equation $3{x^2} + 6x - 2 = 0$.
We have to solve the given quadratic equation by completing the square.
Let us consider the quadratic equation $3{x^2} + 6x - 2 = 0$ ……(1)
First of all to make the coefficient of ${x^2} = 1$, we divide the whole equation by 3 and solve it further.
Dividing the equation given in (1) by 3, we get,
$\dfrac{1}{3}(3{x^2} + 6x - 2) = \dfrac{1}{3} \cdot 0$
$ \Rightarrow \dfrac{3}{3}{x^2} + \dfrac{6}{3}x - \dfrac{2}{3} = 0$
$ \Rightarrow {x^2} + 2x - \dfrac{2}{3} = 0$
Take $\dfrac{2}{3}$ to the R.H.S. and simplify the equation.
Remember that when transferring any variable or number to the other side, the signs of the same will be changed to the opposite sign.
Taking $\dfrac{2}{3}$ to the other side we get,
$ \Rightarrow {x^2} + 2x = \dfrac{2}{3}$ ……(2)
Now we add the square of half the coefficient of x term on both sides of the above equation to complete it as a square and we solve for the variable x.
So in this case adding the ${\left( {\dfrac{2}{2}} \right)^2}$ to both sides of the equation (2), we get,
$ \Rightarrow {x^2} + 2x + {\left( {\dfrac{2}{2}} \right)^2} = \dfrac{2}{3} + {\left( {\dfrac{2}{2}} \right)^2}$
$ \Rightarrow {x^2} + 2x + {1^2} = \dfrac{2}{3} + {1^2}$
$ \Rightarrow {x^2} + 2x + 1 = \dfrac{2}{3} + 1$
Taking LCM on the R.H.S. we get,
$ \Rightarrow {x^2} + 2x + 1 = \dfrac{{2 + 3}}{5}$
$ \Rightarrow {x^2} + 2x + 1 = \dfrac{5}{3}$ ……(3)
Note that the L.H.S. of the above equation is of the form ${a^2} + 2ab + {b^2}$.
We have the algebraic formula given by ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
Here $a = x,$ $b = 1$.
Therefore ${x^2} + 2x + 1$ becomes,
${x^2} + 2x + 1 = {\left( {x + 1} \right)^2}$
Hence substituting this value in the equation (3), we get,
$ \Rightarrow {\left( {x + 1} \right)^2} = \dfrac{5}{3}$
Taking square root on both sides of the equation we get,
$ \Rightarrow x + 1 = \pm \sqrt {\dfrac{5}{3}} $
$ \Rightarrow x = 1 \pm \sqrt {\dfrac{5}{3}} $.

Hence the solution of $3{x^2} + 6x - 2 = 0$ by completing the square is $x = 1 \pm \sqrt {\dfrac{5}{3}} $.

Note:
Here many students solve the equation by using the quadratic formula given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. But they must keep in mind that they have to solve this kind of question by completing the square.
Method for completing the square is explained below.
To solve the quadratic equation of the form $a{x^2} + bx + c = 0$ by completing the square:
(1) If a, the leading coefficient is not equal to 1, then divide the whole equation by a.
(2) Transform the quadratic equation so that the constant term c is alone on the right hand side.
(3) Add the square of half the coefficient of x term, ${\left( {\dfrac{b}{{2a}}} \right)^2}$ to both sides of the equation.
(4) Factor the left hand side as the square of the binomial.
(5) At last solve for the variable x.