How do you solve \[3{{x}^{2}}+4x+3=0\] by completing the square?

Hint: In this problem, we have to solve and find the value of x by completing the square. To get a complete square, we can first divide each term by the coefficient of \[{{x}^{2}}\] , then we can square the coefficient of x, which is multiplied by \[\dfrac{1}{2}\] and add the term on both sides. We will get a complete square from which we can solve the value of x.

Complete step by step answer:
We know that the given quadratic equation to be solved is,
\[3{{x}^{2}}+4x+3=0\]
We can now divide each term by the coefficient of \[{{x}^{2}}\], we get
\[\Rightarrow {{x}^{2}}+\dfrac{4}{3}x=-1\]
Now we can multiply the coefficient of x by \[\dfrac{1}{2}\] and square the term and add it on both the sides in the above equation, we get
\[\Rightarrow {{x}^{2}}+\dfrac{4}{3}x+\dfrac{4}{9}=-1+\dfrac{4}{9}\]
We can cross multiply the terms on the left-hand side and write the right-hand side in a complete square form, we get
\[\Rightarrow {{\left( x+\dfrac{2}{3} \right)}^{2}}=-\dfrac{5}{9}\]
Now we can take square root on both the left-hand side and the right-hand side of the equation, we get
\[\Rightarrow x+\dfrac{2}{3}=\pm \sqrt{-\dfrac{5}{9}}\]
We know that, if we have negative terms inside the square root, we can write it in a complex format, i.e., as an imaginary part.
\[\Rightarrow x+\dfrac{2}{3}=\pm i\dfrac{\sqrt{5}}{3}\]
Now we can subtract \[\dfrac{2}{3}\] on both sides, we get
\[\Rightarrow x=-\dfrac{2}{3}\pm i\dfrac{\sqrt{5}}{3}\]

Therefore, the value of \[x=-\dfrac{2}{3}\pm i\dfrac{\sqrt{5}}{3}\].

Note: Students make mistakes in finding the complete square for the given equation. we should know that we should multiply the coefficient of x by \[\dfrac{1}{2}\] and square the term and add it on both sides to get a complete square. We should also know if we have negative terms inside the square root, we can write it in a complex format, i.e., as an imaginary part.