Try To Factorize it

Since $\alpha$ and $\beta$ are roots of $ax^2+bx+c = 0$ , it implies that $a(x-\alpha)(x-\beta) = ax^2 + bx+c$ . Then we have:

$\begin{aligned} L & = \lim_{x \to \alpha} \frac {1-\cos(ax^2+bx+c)}{(x-\alpha)^2} \\ & = \lim_{x \to \alpha} \frac {1-\color{#3D99F6}\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to \alpha} \frac {1-\color{#3D99F6}\left(1-\frac {a^2(x-\alpha)^2(x-\beta)^2}{2!} + \frac {a^4(x-\alpha)^4(x-\beta)^4}{4!} - \cdots \right)}{(x-\alpha)^2} \\ & = \lim_{x \to \alpha} \frac {\frac {a^2(x-\alpha)^2(x-\beta)^2}{2!} - \frac {a^4(x-\alpha)^4(x-\beta)^4}{4!} + \frac {a^6(x-\alpha)^6(x-\beta)^6}{6!} - \cdots}{(x-\alpha)^2} \\ & = \lim_{x \to \alpha} \left(\frac {a^2(x-\beta)^2}{2!} - \frac {a^4(x-\alpha)^2(x-\beta)^4}{4!} + \frac {a^6(x-\alpha)^4(x-\beta)^6}{6!} - \cdots\right) \\ & = \boxed{\dfrac {a^2}2 (\alpha-\beta)^2} \end{aligned}$