Factorize $6x^2 -5x -14 = 0$ I'm throwing a bit of a blank on the best way to factor this : 
$$6x^2 -5x -14 = 0$$
I know that I could multiply $6$ by $14$ and then find a pair of factors that add to $-5$ (b), but this feels a bit convoluted and I'm sure that I've gone about this a bit of an easier way before, I just can't remember how... 
Does anyone know the simplest way? 
cheers!

\begin{align}
& 6x^2 - 5x - 14 \\[8pt]
= {} & 6\left( x^2 - \frac 5 6 x \right) - 14 \\[8pt]
= {} & 6\left( x^2 - \frac 5 6 x + \frac{25}{144} \right) - 14 - 6 \left(\frac{25}{144}\right) \quad\text{(completing the square)} \\[8pt]
= {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \frac{361}{144}\right) \\[8pt]
= {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \left(\frac{19}{12}\right)^2 \right) \\[8pt]
= {} & 6\left(x - \frac5{12} - \frac{19}{12}\right)\left(x - \frac5{12} + \frac{19}{12}\right) \quad\text{(factoring a difference of two squares)} \\[8pt]
= {} & 6(x-2)\left(x + \frac 7 6\right) \\[8pt]
= {} & (x-2)(6x+7).
\end{align}
If one says
$$
x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{5\pm\sqrt{5^2-4\cdot6\cdot(-14)}}{2\cdot6} = \frac{5\pm19}{12} = \left(\frac{24}{12}\text{ or }\frac{-14}{12}\right) = \left(2\text{ or }\frac{-7}6\right)
$$
then one knows that $x-2$ and $x-\left(\dfrac{-7}6\right)$ must be factors, and one must have $6$ as the leading coefficient, so one gets the same result.
The usual way to prove that
$$
x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
$$
if $ax^2+bx+c=0$ is by completing the square.
