How do you solve the quadratic equation with complex numbers given \[-6{{x}^{2}}+12x-7=0\] ?

Hint: This question belongs to the topic of quadratic equations. We will solve this equation by the method of completing the square. In solving this question, we are going to use the term iota. We are going to use a formula in this question that is \[{{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}\]. After that, we will see an alternate method to solve this question.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to solve the quadratic equation with complex numbers. The quadratic equation is given as \[-6{{x}^{2}}+12x-7=0\]
We will solve this equation by completing the square.
The equation we have to solve is
\[-6{{x}^{2}}+12x-7=0\]
This equation can also be written as
\[\Rightarrow -6{{x}^{2}}+12x-6-1=0\]
After taking -6 as a common factor from the first three terms, we can write the above equation as
\[\Rightarrow -6\left( {{x}^{2}}-2x+1 \right)-1=0\]
The above equation can also be written as
\[\Rightarrow -6\left( {{x}^{2}}-2x+1 \right)=1\]
The above equation can also be written as
\[\Rightarrow {{x}^{2}}-2x+1=\dfrac{1}{-6}\]
Now, we can see that the left side of equation is a perfect square of (x-1). This can be said by the formula \[{{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}\]. So, we can write the above equation as
\[\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{1}{-6}\]
The above equation can also be written as
\[\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{-1}{6}\]
Squaring root both side of the equation, we get
\[\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}}=\pm \sqrt{\dfrac{-1}{6}}=\pm \dfrac{\sqrt{-1}}{\sqrt{6}}\]
We can write the above equation as
\[\Rightarrow x-1=\pm \dfrac{\sqrt{-1}}{\sqrt{6}}\]
Now, let us know about the term iota. The symbol for the term iota is \[i\] and can be written as \[i=\sqrt{-1}\]
So, we can write the above equation as
\[\Rightarrow x-1=\pm \dfrac{i}{\sqrt{6}}\]
The above equation can also be written as
\[\Rightarrow x=1\pm \dfrac{i}{\sqrt{6}}\]
Now, we have solved the equation. We have got the value of x as
\[x=1+\dfrac{i}{\sqrt{6}}\] and \[x=1-\dfrac{i}{\sqrt{6}}\]

Note: For solving this type of question, we should have a better knowledge in the topic of quadratic equations. We should about iota that \[i=\sqrt{-1}\].
Let us solve this question by an alternate method.
For solving this question by alternate method, we should know that
If a quadratic equation is in the form of \[a{{x}^{2}}+bx+c=0\], then the solving formula for the value of x is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
So, after comparing the equation \[a{{x}^{2}}+bx+c=0\] with \[-6{{x}^{2}}+12x-7=0\], we get that
a=-6, b=12, and c=-7.
So, the value of x will be
\[x=\dfrac{-12\pm \sqrt{{{12}^{2}}-4\times \left( -6 \right)\times \left( -7 \right)}}{2\times \left( -6 \right)}=\dfrac{-12\pm \sqrt{144-168}}{-12}=\dfrac{-12\pm \sqrt{-24}}{-12}\]
This can be written as
\[x=\dfrac{-12\pm \sqrt{4\times 6\times \left( -1 \right)}}{-12}=\dfrac{-12\pm 2\sqrt{6}\times \sqrt{\left( -1 \right)}}{-12}=\dfrac{-12}{-12}\pm \dfrac{2\sqrt{6}\times \sqrt{\left( -1 \right)}}{-12}=1\pm \dfrac{2\sqrt{6}\times \sqrt{\left( -1 \right)}}{-2\times 6}\]
As we know that \[i=\sqrt{-1}\]. So, we can write x as
\[x=1\pm \dfrac{i}{-\sqrt{6}}\]
So, the values of x are \[1-\dfrac{i}{\sqrt{6}}\] and \[1+\dfrac{i}{\sqrt{6}}\]
Hence, we have got the same solution. So, we can use this method too.