The point P$(x$, -4) is 13 units away from Q(7, 8). Find the two possible values of $x$.

$(7-x{)}^2+(8-(-4){)}^2=13^2$, so $(7-x{)}^2+144=169$ and $(7-x{)}^2=25$. Then $7-x=\pm 5$, giving $x=2$ and $x=12$. Writing the $y$-difference as 8 - 4 = 4 instead of $8-(-4)=12$ is the usual sign slip here, and it wrecks the whole equation.