$(\alpha-12) x^{2}+2(\alpha-12) x+2=0$

Here,

$a=(\alpha=12), b=2(\alpha-12) \text { and } c=2$

It is given that the roots of the equation are equal; therefore, we have

$\begin{array}{l}
D=0 \\
\Rightarrow\left(b^{2}-4 a c\right)=0 \\
\Rightarrow\{2(\alpha-12)\}^{2}-4 \times(\alpha-12) \times 2=0 \\
\Rightarrow 4\left(\alpha^{2}-24 \alpha+144\right)-8 \alpha+96=0 \\
\Rightarrow 4 \alpha^{2}-96 \alpha+576-8 \alpha+96=0 \\
\Rightarrow 4 \alpha^{2}-104 \alpha+672=0 \\
\Rightarrow \alpha^{2}-26 \alpha+168=0 \\
\Rightarrow \alpha^{2}-14 \alpha-12 \alpha+168=0 \\
\Rightarrow \alpha(\alpha-14)-12(\alpha-14)=0 \\
\Rightarrow(\alpha-14)(\alpha-12)=0 \\
\therefore \alpha=14 \text { or } \alpha=12
\end{array}$

If the value of $\alpha$ is 12 , the given equation becomes non-quadratic. Therefore, the value of $\alpha$ will be 14 for the equation to have equal roots.