Yes, I did several steps at once, perhaps to give you a little challenge (or just to save myself some writing). An intermediate step might have been this:

A = sqrt[(a+b+c)/2((a+b+c)/2 - a)((a+b+c)/2 - b)((a+b+c)/2 - c)]

= sqrt[(a+b+c)/2 ((a+b+c)-2a)/2 ((a+b+c)-2b)/2 ((a+b+c)-2c)/2]

= ...

Can you see what I did now? I multiplied by 2, in 4 places.

I imagine the authors forgot that s depends on c, and wanted you to just solve the equation itself for c, treating s as an independent variable. That's really unforgivable, in that poor students would be happy with it, but good students would struggle to do it right.

You might find it interesting to pick one of the examples, after finding A, and replace one of the sides with x and try solving for that. You will notice, first, that if you allow yourself to use the knowledge of what s is, then you can solve for x just from the definition of s, not using A at all; and, second, that if you replace s with its expression (e.g. s = (4 + 6 + x)/2), you can simplify Heron's formula and have something *a little* less intimidating to actually solve. (I haven't tried it to see how big a challenge it would be; you would obviously have to try to recall what you have learned in the past, which may be more than you want to do just now.)