121x^3+22x^2+x=0
х (121х^2 + 22х + 1)= 0
х= 0
121х^2 + 22х + 1= 0
Д= 22^2 - 4×121×1= 484 - 484= 0
х= -22/(2×121)= -0.09
Ответ: х= 0; х= -0.09.
1) 3x^2-14x+16=0
D=k^2-ac=(-7)^2-3*16=1
k=b/2=-7
x1=7+1/3=8/3
x2=7-1/3=2
2) 5x^2-16x+3=0
D=(-8)^2-5*3=49
x1=8+7/5=3
x2=8-7/5=0.2
3)4x^2-36x+77=0
D=(-18)^2-4*77=16
x1=18+4/4=5.5
x2=18-14/4=3.5
5х^2-8х-4=0
D=64-4*5*(-4)=64+80=144
x1=(8+12)/10=20/10=2
x2=(8-12)/10=-4/10=-2/5