Set $f(x) = \left|20|x| - x^2\right|,$ an even function; the equation says $f(x) = c + 21$ or $f(x) = c - 21.$ For $x \ge 0,$ the graph of $f$ rises from $0$ to $100$ on $[0, 10],$ falls back to $0$ at $x = 20,$ then increases without bound. So for $k$ with $0 \lt k \lt 100,$ the equation $f(x) = k$ has $3$ positive solutions, hence $6$ solutions in all; for $k = 100$ it has $4;$ for $k \gt 100$ it has $2;$ and for $k = 0$ it has $3$ (namely $0$ and $\pm 20$).

The two levels $c + 21$ and $c - 21$ are distinct, so the only way to reach $12$ solutions is $6 + 6:$ both $c - 21$ and $c + 21$ must lie strictly between $0$ and $100.$ This means $c \ge 22$ and $c \le 78,$ and every such integer works: there are $78 - 22 + 1 = 57$ values.