Odpowiedź:
[tex]a)\\\\x^2-7x+10=0\\\\\Delta=b^2-4ac=(-7)^2-4*1*10=49-40=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{7-3}{2}=\frac{4}{2}=2\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{7+3}{2}=\frac{10}{2}=5[/tex]
[tex]b)\\\\2x^2+x-1=0\\\\\Delta=b^2-4ac=1^2-4*2*(-1)=1+8=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-1-3}{2*2}=\frac{-4}{4}=-1\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-1+3}{4}=\frac{2}{4}=\frac{1}{2}[/tex]
[tex]c)\\\\x^2-6x+9=0\\\\(x+3)^2=0\\\\x+3=0\\\\x=-3\\\\\\d)\\\\x^2-25=0\\\\(x-5)(x+5)=0\\\\x-5=0\ \ \ \ \vee\ \ \ \ x+5=0\\\\x=5\ \ \ \ \ \ \ \ \vee\ \ \ \ x=-5[/tex]
[tex]e)\\\\x^2+1=0\\\\x^2=-1\\\\R\'ownanie\ \ nie\ \ ma\ \ rozwiazania\\\\\\f)\\\\x^2=6x\\\\x^2-6x=0\\\\x(x-6)=0\\\\x=0\ \ \ \ \vee\ \ \ \ x-6=0\\\\x=0\ \ \ \ \vee\ \ \ \ x=6[/tex]
