Let $\alpha$ and $\beta$ be the distinct roots of $a{x}^2+bx+c=0$, then ${\lim }_{x\to \alpha }\frac{1-\cos (a{x}^2+bx+c)}{(x-\alpha {)}^2}$ is equal to:

Since $\alpha ,\beta$ are roots of $a{x}^2+bx+c$, we can write:$a{x}^2+bx+c=a(x-\alpha )(x-\beta )$.The limit is:
$$L=\underset{x\to \alpha }{\lim }\frac{1-\cos [a(x-\alpha )(x-\beta )]}{(x-\alpha {)}^2}$$
Using the identity $1-\cos \theta =2{\sin }^2(\theta /2)$:
$$L=\underset{x\to \alpha }{\lim }\frac{2{\sin }^2[\frac{a(x-\alpha )(x-\beta )}{2}]}{(x-\alpha {)}^2}$$
Multiply and divide by $[\frac{a(x-\beta )}{2}{]}^2$:
$$L=\underset{x\to \alpha }{\lim }2\cdot {\left(\frac{\sin [\frac{a(x-\alpha )(x-\beta )}{2}]}{\frac{a(x-\alpha )(x-\beta )}{2}}\right)}^2\cdot \frac{a^2(x-\alpha {)}^2(x-\beta {)}^2}{4(x-\alpha {)}^2}$$
As $x\to \alpha$, the term in the parenthesis $\to 1$:
$$L=2\cdot (1{)}^2\cdot \frac{a^2(\alpha -\beta {)}^2}{4}$$
$$L=\frac{a^2}{2}(\alpha -\beta {)}^2$$