12.4 Let $S=(x-R)^{2}+(x+5)^{2}$, where $x$ is a real number. Find the minimum value of $S$.
$$
\begin{aligned}
S & =(x-10)^{2}+(x+5)^{2} \\
& =x^{2}-20 x+100+x^{2}+10 x+25 \\
& =2 x^{2}-10 x+125 \\
& =2\left(x^{2}-5 x\right)+125 \\
& =2(x-2.5)^{2}+125-2 \times 2.5^{2} \\
& =2(x-2.5)^{2}+112.5 \geq 112.5
\end{aligned}
$$

The minimum value of $S$ is 112.5 .
Method 2
$$
S=(10-x)^{2}+(x+5)^{2}
$$

Let $a=10-x, b=x+5$
$a+b=15$, which is a constant
$\therefore a^{2}+b^{2}$ reaches its minimum when $a=b=7.5$
$\therefore$ Minimum $S=7.5^{2}+7.5^{2}=112.5$