Since, $\alpha$ is a root of given equation,
$\begin{aligned} 25 \cos ^2 \theta+5 \cos \theta-12 & =0 \\ \frac{\pi}{2} & < \alpha < \pi\end{aligned}$
Then, $\begin{aligned} D & =b^2-4 a c=(5)^2-4(25)(-12) \\ & =25+1200=1225\end{aligned}$
Now, $\begin{aligned} \alpha & =\frac{-b \pm \sqrt{D}}{2 a}=\frac{-5 \pm \sqrt{1225}}{50} \\ & =\frac{-5 \pm 35}{50}=\frac{3}{5} \text { or } \frac{-4}{5} \\ \Rightarrow \quad & =\cos \alpha=\frac{3}{5} \text { or }-\frac{4}{5}\end{aligned}$
As $\alpha$ lies in IInd quadrant, $\frac{\pi}{2} < \alpha < \pi$
$\begin{aligned} \therefore \quad \cos \alpha & =-V e=-\frac{4}{5} \\ \text { Now, } \sin 2 \alpha & =2 \sin \alpha \cos \alpha=2 \sqrt{1-\cos ^2 \alpha} \cdot \cos \alpha \\ & =2 \sqrt{1-\frac{16}{25}} \cdot\left(-\frac{4}{5}\right)=2 \cdot \frac{3}{5} \cdot\left(-\frac{4}{5}\right) \\ \sin 2 \alpha & =\frac{-24}{25}\end{aligned}$