Yes I can, now that I tried to read this elaborate link.

The point is that once you know [math]\operatorname{var}(x)[/math] you should be able to apply a few basic rules of the variance operator in order to compute [math]\operatorname{var}(5–3x)[/math].

We have, with [math]a[/math] a constant:

• [math]\operatorname{var}(x+a)=\operatorname{var}(x)[/math]
• [math]\operatorname{var}(ax)=a^2 \operatorname{var}(x)[/math]

But first you should compute [math]\operatorname{var}(x)[/math] using the equation, which is actually given to you in the link.