Odpowiedź:
[tex]a)\\\\x^2-7x-8=0\\\\\Delta=b^2-4ac=(-7)^2-4*1*(-8)=49+32=81\\\\\sqrt{\Delta}=\sqrt{81}=9\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*1}=\frac{7-9}{2}=\frac{-2}{2}=-1\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*1}=\frac{7+9}{2}=\frac{16}{2}=8[/tex]
[tex]b)\\\\3x^2-x=0\\\\x(3x-1)=0\\\\x=0\ \ \ \ \vee\ \ \ \ 3x-1=0\\\\x=0\ \ \ \ \vee\ \ \ \ 3x=1\ \ /:3\\\\x=0\ \ \ \ \vee\ \ \ \ x=\frac{1}{3}[/tex]
[tex]c)\\\\5x^2+2=0\\\\\Delta=b^2-4ac=0^2-4*5*2=0-40=-40\\\\\Delta<0\ \ brak\ \ rozwiazania[/tex]
[tex]d)\\\\\frac{1}{4}x^2-x+1=0\ \ /*4\\\\x^2-4x+4=0\\\\\Delta=b^2-4ac=(-4)^2-4*1*4=16-16=0\\\\\sqrt{\Delta}=\sqrt{0}=0\\\\\\x_{0}=\frac{-b}{2a}=\frac{-(-4)}{2*1}=\frac{4}{2}=2[/tex]
[tex]e)\\\\-2x^2+3x-5=0\\\\\Delta=b^2-4ac=3^2-4*(-2)*(-5)=9-40=-31\\\\\Delta<0\ \ brak\ \ rozwiazania[/tex]
[tex]f)\\\\6x^2-x-5=0\\\\\Delta=b^2-4ac=(-1)^2-4*6*(-5)=1+120=121\\\\\sqrt{\Delta}=\sqrt{121}=11\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*6}=\frac{1-11}{12}=\frac{-10}{12}=-\frac{5}{6}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*6}=\frac{1+11}{12}=\frac{12}{12}=1[/tex]
