Example $3 . x$ is a real number, find
$$
y=2 x^{2}-6 x+10
$$

the minimum value.
\[
\begin{aligned}
y= & 2 x^{2}-6 x+10 \\
= & 2\left(x^{2}-3 x+5\right) \\
& =2\left(x-\frac{3}{2}\right)^{2}+\frac{11}{2}
\end{aligned}
\]
Since \( x \) is a real number, \(\quad \therefore\left(x-\frac{3}{2}\right)^{2} \geqslant 0\),
\(\therefore\) when \( x=\frac{3}{2} \), \( y \) has a minimum value of \(\frac{11}{2}\).