Question

решите уравнение 6x^2+x-2=0,
x^2-x-a^2+a+0
(3x+1)^2=(x+2)^2

Answer 1

1.
1) Д=1+48=49=7^2
2) х1=-1-7/12=-8/12=-2/3
х2=-1+7/12=0.5

2.(х^2-а^2)-(х-а)=0
(х-а)(х+а)-(х-а)=0
х+а=0

3.9х^2+6х+1-х^2-4х-4=0
8х^2+2х-3=0
Д=4+96=100=10^2
х1=-2-10/16=-12/16=-0.75
х2=-2+10/16=8/16=0.5