`a) (x+2)^2+2(x+1)=12`
`⇔ x^2+4x+4+2x+2=12`
`⇔ x^2+4x +2x=12-2-4`
`⇔ x^2+6x-6=0`
`⇔ x^2+6x+9-9-6=0`
`⇔ (x+3)^2-15=0`
`⇔ (x+3)^2=15`
`⇔x+3=± \sqrt15`
`⇔x=-3±\sqrt15 `
Vậy` S={-3+\sqrt15 ;-3-\sqrt15 }`
`b) (x+2)^2-(x-2) (x+1)=3`
`⇔x^2+4x+4-(x^2+x-2x-2)-3=0`
`⇔x^2+4x+4-x^2-x+2x+2-3=0`
`⇔ 5x=-3`
`⇔x=\frac{-3}{5}`
Vậy `S={`$\dfrac{-3}{5}$`}`