<s>[INST] if $mx+3|x+4|-2=0$ has no solutions, solve for $m$ 
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$

my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefore
$$(4m+72)^2-4(9-m^2)140<0 \\ 576m^2+576m+144<0 \\ 4m^2+4m+1<0 \\ (2m+1)^2<0$$
maybe I made a mistake but I couldn't find it
 [/INST] The hint.
For $x\geq-4$ we obtain:
$$mx+3x+12-2=0,$$ which gives a value $m=-3$.
For $x\leq-4$ we obtain $$mx-3x-12-2=0,$$ which gives a value $m=3.$
Now, check that for $m=3$ our equation has root, while for $m=-3$ our equation has no roots.
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