`x²-5x-35=0`
`⇔x²-5x+25/4-165/4=0`
`⇔[x²-2*x*2/5+(5/2)^2]-((\sqrt{165)/2)²=0`
`⇔(x-5/2)²-((\sqrt{165)/2)²=0`
`⇔(x-5/2-(\sqrt{165)/2)(x-5/2+(\sqrt{165)/2)=0`
`1) x-5/2-\sqrt{165)/2=0⇔x=(5+\sqrt{165})/2`
`2) x-5/2+\sqrt{165)/2=0⇔x=(5-\sqrt{165})/2`
Vậy `S={(5±\sqrt{165})/2}`
`b) x²-4=(3x+1)(x+2)`
`⇔(x+2)(x-2)-(3x+1)(x+2)=0`
`⇔(x+2)(x-2-3x-1)=0`
`⇔(x+2)(-2x-3)=0`
`1) x+2=0⇔x=-2`
`2) -2x-3=0⇔x=-3/2`
Vậy `S={-2;-3/2}`
