Odpowiedź:
[tex]a)\\\\3x^2-10x+3 < 0\\\\a=3\ \ ,\ \ b=-10\ \ ,\ \ c=3\\\\\Delta=b^2-4ac\\\\\Delta=(-10)^2-4\cdot3\cdot3=100-36=64\\\\\sqrt{\Delta}=\sqrt{64}=8\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8}{2\cdot3}=\frac{10-8}{6}=\frac{2}{6}=\frac{1}{3}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8}{2\cdot3}=\frac{10+8}{6}=\frac{18}{6}=3[/tex]
[tex]b)\\\\x^2-3x+2 > 0\\\\a=1\ \ ,\ \ b=-3\ \ ,\ \ c=2\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot1\cdot2=9-8=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2\cdot1}=\frac{3-1}{2}=\frac{2}{2}=1\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2\cdot1}=\frac{3+1}{2}=\frac{4}{2}=2[/tex]
