Algebraic Mod

We have $(x+12)^2-144+31\equiv 0$ so $(x+12)^2\equiv 113\equiv 16=4^2$ and $x\equiv \pm4-12$ . The representatives on $[0,100]$ are $97-4-12=81$ and $97+4-12=89$ , and the answer is $\boxed{170}$