That is, (-m)2-4(2m- 1)=m2-8m+4≥0,

The solution is the root number 3 of m≥4+2.

Or m≤4? 2 root number 3

Let the two roots of the original equation be α and β, then α+β=m and α β = 2m- 1.

α2+β2=α2+β2+2αβ-2αβ

=(α+β)2-2αβ

=m2-2(2m- 1)

=m2-4m+2=7。

That is, m2-4m-5 = 0.

The solution is m=- 1 or m=5.

∫m = 5≤4+2 Radical 3

, ∴m=5 (give up)

∴ m =- 1。 So choose B.

I hope it helps you!