Odpowiedź:
[tex]a)\\\\2x^2-3x+1<0\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot2\cdot1=9-8=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2\cdot2}=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2\cdot2}=\frac{3+1}{4}=\frac{4}{4}=1[/tex]
[tex]b)\\\\-(x+1)(x-5)\geq 0\ \ /\cdot(-1)\\\\(x+1)(x-5)\leq 0\\\\x+1=0\ \ \ \ \vee\ \ \ \ x-5=0\\\\x=-1\ \ \ \ \ \ \vee\ \ \ \ x=5[/tex]
