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                                                вит39 
                                                14.07.2019 19:40
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                                                raxmatullin174 
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                                                14.07.2019 19:40
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                                                Filipok220 
                                                14.07.2019 19:40
                                            
2x2−x=0
x(2x−1)=0()
x=0 or 2x−1=0
x=0 or x=12
x<0
0<x<12)
x>12
Answer:
0<x<12
ХЭ(0;12)
x2+3x<0
x2+3x=0
x(x+3)=0
x=0 or x+3=0
x=0 or x=−3
Check intervals in between critical points. (Test values in the intervals to see if they work.)
x<−3
−3<x<0
x>0(
Answer:
−3<x<0
ХЭ(-3;0)