Question 869916
{{{(x+2)(2x+1)<=0}}}
Break up the number line into 3 regions.
{{{x<-2}}}
{{{-2<x<-1/2}}}
{{{x>-1/2}}}
Test the inequality in each region with a point. 
{{{x<-2}}}, {{{x=-3}}}, {{{(-3+2)(2(-3)+1)<=0}}},{{{(-1)(-5)<=0}}},{{{5<=0}}} False
{{{-2<x<-1/2}}},{{{x=-1}}},{{{(-1+2)(2(-1)+1)<=0}}},{{{(1)(-1)<=0}}},{{{-1<=0}}} True
{{{x>-1/2}}},{{{x=0}}},{{{(0+2)(2(0)+1)<=0}}},{{{2<=0}}}, False
So the only region where the inequality is true is, 
{{{highlight(-2<=x<=-1/2)}}}