Solution:
By the Cauchy-Schwarz Inequality,
$$[2(x-1)+4(2 y)]^{2} \leq (2^{2}+4^{2})\left[(x-1)^{2}+4 y^{2}\right]$$
Now, with $2(x-1)+4(2 y)=2 x+8 y-2=3-2=1$, we obtain
$$\begin{aligned} x^{2}+4 y^{2}-2 x & =(x-1)^{2}+4 y^{2}-1 \\ & \geq \frac{[2(x-1)+4(2 y)]^{2}}{2^{2}+4^{2}}-1 \\ & =\frac{1}{20}-1 \\ & =-\frac{19}{20} \end{aligned}$$
The minimum $-\frac{19}{20}$ is indeed obtained with $x=\frac{11}{10}$ and $y=\frac{1}{10}$.