=>(x+1).(2-x)-3x-5 +4x2-2=0
=>(x+1).(2-x)+(4x2-3x-7)=0
=>(x+1).(2-x)+(4x2 -7x+4x-7)=0
=>(x+1)(2-x)+[x.(4x-7)+(4x-7)]=0
=>(x+1)(2-x)+(4x-7).(x+1)=0
=>(x+1).[(2-x)+(4x-7)]=0
=>(x+1).(2-x+4x-7)=0
=>(x+1).(3x-5)=0
=>hoặc x+1=0 hoặc 3x-5=0
=>hoặc x=-1 hoặc x=5/3