7. Given that the sum of the squares of the two real roots of the equation $2 x^{2}+k x-2 k+1=0$ is $\frac{29}{4}$, then the value of $k$ is
(A) 3 ;
(B) -11 ;
(C) 3 or -11 ；
(D) 11 ;
(E) None of the above conclusions are correct.
7. (A)

Since the equation has two real roots, then
$$
\Delta=k^{2}-4 \cdot 2(-2 k+1)>0 .
$$

Given that the sum of the squares of the roots is $\frac{29}{4}$, then
$$
x_{1}^{2}+x_{2}^{2}=\frac{29}{4} \text {. }
$$

From (1), we have
$$
k^{2}+16 k-8>0,
$$

Solving this gives $k > -8 + 6 \sqrt{2}$.
From (2), we have
$$
\begin{aligned}
x_{1}^{2}+x_{2}^{2} & =\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
& =\left(-\frac{k}{2}\right)^{2}-2 \frac{-2 k+1}{2}=\frac{29}{4},
\end{aligned}
$$

which simplifies to $k^{2}+8 k-33=0$.
Solving this gives $k=3$ or $k=-11$.
However, $k=-11$ does not satisfy the range in (3), so $k=3$ is the solution.