= Solution
For the <Dahlquist test equation>, write $z=h\lambda$. The amplification roots solve
$$
(7-4z)\zeta^2-12\zeta+5+2z=0.
$$
A root on $|\zeta|=1$ would require the boundary-locus value
$$
z=\frac{7\zeta^2-12\zeta+5}{4\zeta^2-2}.
$$
Its denominator cannot vanish on that circle. Setting $\zeta=e^{i\theta}$ and simplifying the real part gives
$$
\operatorname{Re}z=\frac{3(1-\cos\theta)^2}{9-8\cos^2\theta}\geq0.
$$
Thus no amplification root can cross the unit circle while $\operatorname{Re}z<0$. The leading coefficient is also nonzero throughout that half-plane. At $z=-1$, the roots are $(6\pm\sqrt3)/11$, both strictly inside the unit disk. Connect any other left-half-plane point to $-1$: continuity of the roots, without a boundary crossing or a root escaping through infinity, keeps both roots inside.
On the imaginary axis a boundary root can occur only at $\theta=0$, which gives $z=0$; there the roots $1$ and $5/7$ satisfy the simple-unit-root condition. Hence \b[$\boxed{\text{the method is A-stable}}$], including the usual boundary root convention at zero.
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