Odpowiedź:
[tex]\frac{1}{6}x^2-1\frac{1}{3}x=-2\\\\\frac{1}{6}x^2-\frac{4}{3}x=-2\ \ /\cdot6\\\\x^2-8x=-12\\\\x^2-8x+12=0\\\\x^2-2x-6x+12=0\\\\x(x-2)-6(x-2)=0\\\\(x-2)(x-6)=0\\\\x-2=0\ \ \ \ \vee\ \ \ \ x-6=0\\\\x=2\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=6[/tex]
II sposób
[tex]\frac{1}{6}x^2-1\frac{1}{3}x=-2\\\\\frac{1}{6}x^2-\frac{4}{3}x=-2\ \ /\cdot6\\\\x^2-8x=-12\\\\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{8-4}{2}=\frac{4}{2}=2\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2\cdot1}=\frac{8+4}{2}=\frac{12}{2}=6[/tex]
