To find the equation of a straight line with a gradient (slope) of $-2$ that passes through the point $(4,1)$, we use the point-gradient form of the line equation:

$$y-y_1=m(x-x_1)$$

where:
- $m$ is the gradient,
- $(x_1,y_1)$ is the point the line passes through.

For the given line:
- $m=-2$,
- $(x_1,y_1)=(4,1)$.

Substitute these values into the point-gradient form:

$$y-1=-2(x-4)$$

Expand and simplify:

$$y-1=-2x+8$$

$$y=-2x+8+1$$

$$y=-2x+9$$

To put this in the standard form $Ax+By+C=0$, rearrange:

$$2x+y-9=0$$

Conclusion:
The correct equation of the line is C. $y+2x-9=0$.