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**Hint:**The given equation is in the $a{{x}^{2}}+bx+c=0$ from. If the equation is in standard form so simplify it replace the $'y'$ with $'O'$ We have to do this because the roots of equation are the value where $y$-axis equal to $0.$ We have the formula for finding the roots substitute the given value in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for the roots. By using the given method, solve the problem.

**Complete step by step solution:**

We have given equation,

${{x}^{2}}-6x+12=0...(i)$

i.e. $0={{x}^{2}}-6x+12$

The above equation is equating to $0.$

So, we must find a solution that is for $y=0$

Consider,

$y=a{{x}^{2}}+bx+c$ and $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\therefore a=1,b=-6,c=12$

We know that the equation part ${{b}^{2}}-4ac$ is known as determined.

By solving,

$\Rightarrow \because {{b}^{2}}-4ac={{\left( -6 \right)}^{2}}-4\left( 1 \right)\times \left( 12 \right)$

$\therefore {{b}^{2}}-4ac=-12$

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow x=\dfrac{-\left( -6 \right)\pm \sqrt{-12}}{2\times 1}$ $\left( {{b}^{2}}-4ac=-12 \right)$

$\Rightarrow x=\dfrac{+6\pm \sqrt{-12}}{2}$

$\Rightarrow x=3\pm \sqrt{\dfrac{-12}{4}}$

$\Rightarrow $\[x=3\pm \sqrt{3\times \left( -1 \right)}\]

$\Rightarrow x=3\pm \sqrt{3}\times \sqrt{-1}$

But, $\sqrt{-1}=i$

The,

$x=3\pm \sqrt{3}i$

**Hence,The roots for the equation ${{x}^{2}}-6x+12=0$ is $x=3\pm \sqrt{3}i$**

**Additional Information:**

A quadratic equation can have one two or no real solution, which are actually the roots of the equation, Which are the points where the equation crosses the $x$-axis. Solving the quadratic equation for its roots and there is one more method for solving by the quadratic formula. For getting the root you use various methods but confirm whether it is correct or not. Writer quadratic equation for example ${{x}^{2}}+3x+2=0$ and for this equation substitute $-1$ into the equation which result us $0=0$ we can say that the first root is correct. Now substitute $-2$ into the equation which results in $0=0.$ Since we got $0$ by solving both the roots. Hence roots for the given equation is $-1\And -2.$

**Note:**Compare the given equation is in the standard form of $a{{x}^{2}}+b+c=0$ and simplify or solving for $x$ we have the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Remember this after finding the roots they are in both cases such the roots are negative and positive. For solving easily we have to solve first ${{b}^{2}}-4ac$ term for simplifying it easily.

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