Odpowiedź:
[tex]a)\\\\5x^2-4x-1=0\\\\a=5\ \ ,\ \ b=-4\ \ ,\ \ \ c=-1\\\\\\\Delta=b^2-4ac=(-4)^2-4*5*(-1)=16+20=36\\\\\sqrt{\Delta}=\sqrt{36}=6\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*5}=\frac{4-6}{10}=\frac{-2}{10}=-\frac{1}{5}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*5}=\frac{4+6}{10}=\frac{10}{10}=1[/tex]
[tex]b)\\\\x^2-5x+6=0\\\\a=1\ \ ,\ \ b=-5\ \ ,\ \ c=6\\\\\\\Delta=b^2-4ac=(-5)^2-4*1*6=25-24=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{5-1}{2}=\frac{4}{2}=2\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{5+1}{2}=\frac{6}{2}=3[/tex]
