21a^2 * ((a^3-4a+2)/7)*4a*((a^4-2a^2+2)/0, 2)+a^3+b

ответ:

d=b^2-4ac=(-1)^2-4*1*(-72)=1+288=\sqrt{289}

289

=17

х1=\frac{-b- \sqrt{d} }{2a} = \frac{1-17}{2} = \frac{-16}{2} =-8

2a

−b−

d

=

2

1−17

=

2

−16

=−8

х2=\frac{-b+ \sqrt{d} }{2a} = \frac{1+17}{2} = \frac{18}{2} = 9

2a

−b+

d

=

2

1+17

=

2

18

=9

ответ: -8 и 9

d=b^2-4ac=7^2-4*(-4)*(-3)=49-48=\sqrt{1} =1

1

=1

х1=\frac{-b- \sqrt{d} }{2a} = \frac{-7-1}{2*(-4)} = \frac{-8}{-8} =1

2a

−b−

d

=

2∗(−4)

−7−1

=

−8

−8

=1

х2=\frac{-b+ \sqrt{d} }{2a} = \frac{-7+1}{(-8)} = \frac{-6}{-8} =0,75

2a

−b+

d

=

(−8)

−7+1

=

−8

−6

=0,75