Odpowiedź:
[tex]a)\\\\y=-4x^2+3x+1=0\\\\a=-4\ \ ,\ \ b=3\ \ \ \ ,\ \ c=1\\\\\Delta=b^2-4ac\\\\\Delta=3^2-4\cdot(-4)\cdot1=9+16=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-3-5}{2\cdot(-4)}=\frac{-8}{-8}=1\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-3+5}{2\cdot(-4)}=\frac{2}{-8}=-\frac{1}{4}\\\\\\y=a(x-x_{1})(x-x_{2})\\\\y=-4(x-1)(x-(-\frac{1}{4}))\\\\y=-4(x-1)(x+\frac{1}{4})[/tex]
[tex]b)\\\\y=2x^2-3x+4=0\\\\a=2\ \ \ \ ,\ \ \ \ b=-3\ \ \ \ ,\ \ \ \ c=4\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot2\cdot4=9-32=-23\\\\\Delta < 0\ \ \ \ posta\'c\ \ iloczynowa\ \ nie\ \ istnieje\ \ (brak\ \ miejsc\ \ zerowych)[/tex]
[tex]c)\\\\y=x^2-2x-2=0\\\\a=1\ \ \ \ ,\ \ \ \ b=-2\ \ \ \ ,\ \ \ \ c=-2\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot1\cdot(-2)=4+8=12\\\\\sqrt{\Delta}=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2\cdot1}=\frac{2-2\sqrt{3}}{2}=\frac{\not2(1-\sqrt{3})}{\not2}=1-\sqrt{3}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2\cdot1}=\frac{2+2\sqrt{3}}{2}=\frac{\not2(1+\sqrt{3})}{\not2}=1+\sqrt{3}[/tex]
[tex]y=a(x-x_{1})(x-x_{2})\\\\y=1(x-(1-\sqrt{3}))(x-(1+\sqrt{3}))\\\\y=(x-1+\sqrt{3})(x-1-\sqrt{3})[/tex]
[tex]d)\\\\y=2x^2-2x-1\\\\a=2\ \ \ \ ,\ \ \ \ b=-2\ \ \ \ ,\ \ \ \ c=-1\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot2\cdot(-1)=4+8=12\\\\\sqrt{\Delta}=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2\cdot2}=\frac{2-2\sqrt{3}}{4}=\frac{\not2^1(1-\sqrt{3})}{\not4_{2}}=\frac{1-\sqrt{3}}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2\cdot2}=\frac{2+2\sqrt{3}}{4}=\frac{\not2^1(1+\sqrt{3})}{\not4_{2}}=\frac{1+\sqrt{3}}{2}[/tex]
[tex]y=a(x-x_{1})(x-x_{2})\\\\y=2(x-\frac{1-\sqrt{3}}{2})(x-\frac{1+\sqrt{3}}{2})[/tex]
