How do you factor $2{{x}^{2}}-x+6$?

Hint: Now we know that the given equation is a quadratic equation. We can easily find the roots of the equation by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Now we know that if $\alpha $ and $\beta $ are the roots of the equation then $x-\alpha $ and $x-\beta $ are the factors of the equation.

Complete step-by-step solution:
Now consider the given equation $2{{x}^{2}}-x+6$ .
The given equation is a quadratic equation in one variable of the form $a{{x}^{2}}+bx+c=0$ where a = 2, b = -1 and c = 6. Now we know that the roots of quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now substituting the values of a, b and c we will get the roots of the given equation.
Hence we have $x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 2 \right)\left( 6 \right)}}{2a}$
$\Rightarrow x=\dfrac{1\pm \sqrt{1-48}}{2\left( 2 \right)}$
$\Rightarrow x=\dfrac{1\pm \sqrt{47}i}{4}$
Hence the factors of the given equation are $\dfrac{1+\sqrt{47}i}{4}$ and $\dfrac{1-\sqrt{47}i}{4}$.
Now we know that if $\alpha $ and $\beta $ are the roots of the equation then $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ are the factors of the equation.
Hence we have $\left( x-\dfrac{1+\sqrt{47}i}{4} \right)$ and $\left( x-\dfrac{1-\sqrt{47}i}{4} \right)$ are the factors of the given equation.
Hence we have $\left( x-\dfrac{1+\sqrt{47}i}{4} \right)\left( x-\dfrac{1-\sqrt{47}i}{4} \right)=2{{x}^{2}}-x+6$.

Note: Now note that the nature of the roots of the equation is given by discriminant which is given by $D={{b}^{2}}-4ac$ . If D > 0 then the roots of the equation are real and distinct. If D = 0 then the roots of the equation are real and equal and if D < 0 then the roots of the equation are complex roots. In the given example the roots are complex since discriminant is negative. Note that the complex roots are always conjugate to each other.