Odpowiedź:
[tex](x+6)(x+2)=0\\\\x^2+2x+6x+12=0\\\\x^2+8x+12=0\\\\a=1\ \ ,\ \ b=8\ \ ,\ \ c=12\\\\\Delta=b^2-4ac\\\\\Delta=8^2-4\cdot1\cdot12=64-48=16\\\\\sqrt{\Delta}=\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-8-4}{2\cdot1}=\frac{-12}{2}=-6\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-8+4}{2\cdot1}=\frac{-4}{2}=-2[/tex]
