Odpowiedź :
[tex]x_1+x_2=-\dfrac{b}{a}\\x_1x_2=\dfrac{c}{a}\\\\\\x^2-9x-7=0\\\\\\x_1+x_2=-\dfrac{-9}{1}=9\\x_1x_2=\dfrac{-7}{1}=-7[/tex]
[tex]x^{2} -9x-7=0\\\\a=1,~~b=-9,~~c=-7\\\\\Delta=x^{2} -4\cdot a\cdot c\\\\\Delta =(-9)^{2} -4\cdot 1\cdot (-7)\\\\\Delta = 81+28=109\\\\\sqrt{\Delta } =\sqrt{109} \\\\x_{1}=\dfrac{-b-\sqrt{\Delta} }{2\cdot a} ~~\lor~~x_{2}=\dfrac{-b+\sqrt{\Delta} }{2\cdot a}\\\\\\x_{1}=\dfrac{9-\sqrt{109} }{2}~~\lor ~~x_{2}=\dfrac{9+\sqrt{109} }{2}\\\\\\x_{1} +x_{2} =\dfrac{9-\sqrt{109} }{2}+\dfrac{9+\sqrt{109} }{2}\\\\\\x_{1} +x_{2} =\dfrac{9-\sqrt{109} +9+\sqrt{109} }{2}[/tex]
[tex]x_{1} +x_{2}=\dfrac{18}{2} \\\\x_{1} +x_{2}=9[/tex]
[tex]x_{1} \cdot x_{2} =\dfrac{9-\sqrt{109} }{2}\cdot \dfrac{9+\sqrt{109} }{2}\\\\\\x_{1} \cdot x_{2} =\dfrac{(9-\sqrt{109} )\cdot (9+\sqrt{109} ) }{2\cdot 2}\\\\\\x_{1} \cdot x_{2}=\dfrac{9^{2} -(\sqrt{109} )^{2} }{4} \\\\\\x_{1} \cdot x_{2}=\dfrac{81-109}{4} \\\\\\x_{1} \cdot x_{2}=\dfrac{-28}{4} \\\\\\x_{1} \cdot x_{2}=-7[/tex]