`\qquad x^2+(m-1)x-3m+7=0`
`\Delta=(m-1)^2-4(-3m+7)`
`\Delta=m^2-2m+1+12m-28`
`\Delta=m^2+10m-27`
Để pt có 2 nghiệm phân biệt
`<=> \Delta>0`
`=> m^2+10m-27>0`
`<=> (m^2+10m+25)-52>0`
`<=> (m+5)^2-52>0`
`<=> (m+5+2\sqrt{13})(m+5-2\sqrt{13})>0`
`<=> [({(m+5+2\sqrt{3}>0),(m+5-2\sqrt{13}>0):}),({(m+5+2\sqrt{3}<0),(m+5-2\sqrt{13}<0):}):}`
`<=> [({(m> -5-2\sqrt{3}),(m> -5+2\sqrt{13}):}),({(m<-5-2\sqrt{3}),(m<-5+2\sqrt{13}):}):}`
`<=> [(m> -5+2\sqrt{13}),(m<-5-2\sqrt{13}):}`
Vậy `m∈(-oo;-5-2\sqrt{13})∪(-5+2\sqrt{13};+oo)`
