Question

Помогите, пожалуйста...

Answer 1

$D=b^{2}-4ac=(a+c)^{2}-4ac=a^{2}+2ac+c^{2}-4ac=a^{2}-2ac+c^{2}= \ =(a-c)^{2} \ x_{1}= frac{-b+ sqrt{D}}{2a}= frac{-(a+c)+ sqrt{(a-c)^{2}}}{2a}= frac{-a-c+a-c}{2a}= frac{-2c}{2a}=- frac{c}{a} \x_{2}= frac{-b- sqrt{D}}{2a}= frac{-(a+c)- sqrt{(a-c)^{2}}}{2a}= frac{-a-c-a+c}{2a}= frac{-2a}{2a}=-1$