Verified answer

Ответ:

[tex]x^2+x=0\\x^2+x+0=0\\[/tex]

т.е a = 1; b = 1; c = 0

D = [tex]b^2-4ac=1-0=1[/tex]

[tex]x_{1}=\frac{-b+\sqrt{D} }{2a}=\frac{-1+1}{2} = 0\\x_{2} = \frac{-b-\sqrt{D} }{2a}=\frac{-1-1}{2} = -\frac{2}{2}=-1[/tex]

Конечный ответ: 0 и -1

[tex]\displaystyle\bf\\x^{2} +x=0\\\\x\cdot(x+1)=0\\\\\\\left[\begin{array}{ccc}x=0\\x+1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =0\\x_{2} =-1\end{array}\right\\\\\\Otvet \ : \ 0 \ \ ; \ \ -1[/tex]