Odpowiedź:
[tex]a)\ \ y=x^2-6x+5\\\\x^2-6x+5=0\\\\a=1\ \ ,\ \ b=-6\ \ ,\ \ c=5\\\\\Delta=b^2-4ac\\\\\Delta=(-6)^2-4\cdot1\cdot5=36-20=16\\\\\sqrt{\Delta}=\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2\cdot1}=\frac{6-4}{2}=\frac{2}{2}=1\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2\cdot1}=\frac{6+4}{2}=\frac{10}{2}=5\\\\\\y=a(x-x_{1})(x-x_{2})\\\\y=1(x-1)(x-5)\\\\y=(x-1)(x-5)[/tex]
[tex]b)\ \ y=x^2+x-2\\\\x^2+x-2=0\\\\a=1\ \ ,\ \ b=1\ \ ,\ \ c=-2\\\\\Delta=b^2-4ac\\\\\Delta=1^2-4\cdot1\cdot(-2)=1+8=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-1-3}{2\cdot1}=\frac{-4}{2}=-2\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-1+3}{2\cdot1}=\frac{2}{2}=1\\\\\\y=a(x-x_{1})(x-x_{2})\\\\y=1(x-(-2))(x-1)\\\\y=(x+2)(x-1)[/tex]
