Square both sides: $\((x-5)^2=2 x^2-18 x+37\)$.
Then

$$
\(x^2-10 x+25=2 x^2-18 x+37\)
$$

so

$$
\(0=x^2-8 x+12\)
$$

which factors as

$$
\(0=(x-2)(x-6) \cdot\left[\text { code }_f \text { ile }: 0\right]\)
$$


So $x=2$ or $x=6$.
Check the original equation and the domain condition $\(x-5 \geq 0\)$, so $\(x \geq 5\)$.

- For $x=2$ : left side $\(2-5=-3\)$, right side is a square root (nonnegative), so this is invalid.

- For $x=6$ : left side $6-5=1$, right side $\(\sqrt{2 \cdot 36-18 \cdot 6+37}=\sqrt{72-108+37}= \sqrt{1}=1\)$, so it works.

Thus $x$ could equal 6 , which is option E .