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$|3x^2+7x|=4\\ \left [{ {{3x^2+7x=-4} \atop {3x^2+7x=4}} \right. \ \left [{ {{3x^2+7x+4=0} \atop {3x^2+7x-4=0}} \right. \\3x^2+7x+4=0\\D=49-4\cdot 4\cdot 3=1\\x_{1,2}=\frac{-7\pm1}{6} =\left |{ {{-1} \atop {-\frac{4}{3} }} \right. \\3x^2+7x-4=0\\D=49+48=97\\x_{1,2}=\frac{-7\pm\sqrt{97}}{6}=\left [{ {{-\frac{7-\sqrt{97}}{6}} \atop {-\frac{7+\sqrt{97}}{6}}} \right.$

Ответ: $-\frac{4}{3};-1;-\frac{7-\sqrt{97}}{6} ;-\frac{7+\sqrt{97}}{6}$

$x^2-5|x|+4=0\\(|x|)^2-5|x|+4=0\\|x|=t;t\geq 0\\t^2-5t+4=0\\\left [{ {{t=4} \atop {t=1}} \right. \left [{ {{|x|=4} \atop {|x|=1}} \right. \left [{ {{\left [{ {{x=4} \atop {x=-4}} \right. } \atop {\left [{ {{x=-1} \atop {x=1}} \right. }} \right.$

Ответ: $-4;-1;1;4$