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Rewriting the left-hand side, we get $$|x(x-5)|=3.$$Therefore, either $x(x-5)=3$ or $x(x-5)=-3.$ These are equivalent to $x^2-5x-3=0$ and $x^2-5x+3=0,$ respectively. The discriminant of both quadratic equations is positive, so they both have two real roots for $x.$ By Vieta's formulas, the sum of the roots of each quadratic is $5,$ so the sum of all four roots is $5+5=\boxed{10}.$