= Solution
For a closed piecewise smooth path $\gamma$ avoiding $w$, its <winding number> is
$$
\boxed{n(\gamma,w)=\frac1{2\pi i}\int_\gamma\frac{dz}{z-w}}.
$$
A meromorphic function has a zero of order $m>0$ at $z_0$ when
$$
f(z)=(z-z_0)^m g(z),
\qquad g(z_0)\ne0,
$$
with $g$ holomorphic; it has a pole of order $m$ at $w_0$ when $(z-w_0)^mf(z)$ extends holomorphically and nonvanishingly there.
The <argument principle> states that, if $f$ has no zero or pole on $\gamma$,
$$
\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\,dz
=\sum_a n(\gamma,a)\operatorname{ord}_a(f).
$$
For a positively oriented simple contour this is the number of enclosed zeros minus poles, counted with multiplicity. It follows from the <residue theorem> because the residue of $f'/f$ at a zero of order $m$ is $m$, while at a pole of order $m$ it is $-m$.
Let
$$
p(z)=z^4+10z^3+4z^2+10z+5.
$$
On the imaginary axis,
$$
p(iy)=\underbrace{(y^4-4y^2+5)}_{(y^2-2)^2+1>0}
+10iy(1-y^2).
$$
Thus $p(iy)$ always lies in the open right half-plane, has no zero there, and its continuous argument has net change zero as the imaginary-axis part of a large right-half-disc contour is traversed. On the right semicircle, $p(z)/z^4\to1$ uniformly, so the argument change tends to that of $z^4$, namely $4\pi$. The argument principle therefore gives
$$
\boxed{\frac{4\pi}{2\pi}=2}
$$
roots in the open right half-plane, counted with multiplicity.
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