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

Hint: We first make the coefficient of \[{x^2}\] as 1 by dividing the complete equation by the coefficient. Then shift the constant value to the right hand side of the equation. Add the square of half value of coefficient of x on both sides of the equation. Use the identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] to write the left side of the equation. Take the square root on both sides of the equation and write the value by solving the equation. Take LCM on the left side and cross multiply values.

Complete step-by-step solution:
We are given the quadratic equation \[3{x^2} - 2x - 3 = 0\]
We compare the equation with general quadratic equation i.e. \[a{x^2} + bx + c = 0\]
Coefficient of \[{x^2}\] is 3, coefficient of x is -2 and constant value is -3.
Now we have to make coefficient of \[{x^2}\] as 1, so we divide the complete equation by 3
\[ \Rightarrow \dfrac{{3{x^2} - 2x - 3}}{3} = 0\]
\[ \Rightarrow \dfrac{{3{x^2}}}{3} - \dfrac{{2x}}{3} - \dfrac{3}{3} = 0\]
Cancel same factors from numerator and denominator
\[ \Rightarrow {x^2} - \dfrac{2}{3}x - 1 = 0\]
Now we shift the constant value to right hand side of the equation
\[ \Rightarrow {x^2} - \dfrac{2}{3}x = 1\]
The coefficient of x is \[ - \dfrac{2}{3}\], we divide the number by 2 and square it i.e. \[{\left( { - \dfrac{2}{{3 \times 2}}} \right)^2} = \dfrac{1}{9}\]
Add this term to both sides of the equation
\[ \Rightarrow {x^2} - \dfrac{2}{3}x + \dfrac{1}{9} = 1 + \dfrac{1}{9}\]
We can write left hand side of the equation using the identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] and simplify right hand side of the equation by taking LCM
\[ \Rightarrow {\left( {x - \dfrac{1}{3}} \right)^2} = \dfrac{{9 + 1}}{9}\]
\[ \Rightarrow {\left( {x - \dfrac{1}{3}} \right)^2} = \dfrac{{10}}{{{3^2}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{{\left( {x - \dfrac{1}{3}} \right)}^2}} = \sqrt {\dfrac{{10}}{{{3^2}}}} \]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow \left( {x - \dfrac{1}{3}} \right) = \pm \dfrac{{\sqrt {10} }}{3}\]
Now equate left side with right side of the equation
\[ \Rightarrow x - \dfrac{1}{3} = \dfrac{{\sqrt {10} }}{3}\]and \[x - \dfrac{1}{3} = - \dfrac{{\sqrt {10} }}{3}\]
Shift constant values to right side
\[ \Rightarrow x = \dfrac{{\sqrt {10} }}{3} + \dfrac{1}{3}\]and \[x = - \dfrac{{\sqrt {10} }}{3} + \dfrac{1}{3}\]
Since denominator is same for both fractions, we can add numerators
\[ \Rightarrow x = \dfrac{{1 + \sqrt {10} }}{3}\] and \[x = \dfrac{{1 - \sqrt {10} }}{3}\]

\[\therefore \]The solution of \[3{x^2} - 2x - 3 = 0\] is \[x = \dfrac{{1 + \sqrt {10} }}{3}\] and \[x = \dfrac{{1 - \sqrt {10} }}{3}\]

Note: Many students make mistakes in starting of the question as they don’t make the coefficient of \[{x^2}\] as 1 and begin by shifting constant to RHS which is wrong. This will make the whole solution incorrect. Keep in mind this solution will be correct if the quadratic equation is in standard form first. Also, sometimes students square the value of b and then divide by 2 instead they have to first divide by 2 and then square it.