Помогите решить

1) 3 х³+8 х²-15 х-40=0

2) (х²+2 х) ²-10 (х²+2 х) + 21=0

1) 3x³+8x²-15x-40=0

x² (3x+8) - 5 (3x+8) = 0

(3x+8) (x²-5) = 0

a) 3x+8=0, x=-8/3

b) x²-5=0, (x+√5) (x-√5) = 0, x = - √5, x=+√5

Otvet: x1=-8/3, x2=-√5, x3=√5

2) (x²+2x) ²-10 (x²+2x) + 21=0

(x²+2x) = z

z²-10z+21=0, D=100-84=16, √D=√16=4

z1 = (10+4) / 2=14/2=7, z1=7

z2 = (10-4) / 2=6/2=3, z2=3

a) x²+2x=7, x²+2x-7=0, D=4+28=32=16.2,√ (16.2) = 4.√2

x1 = (-2+4√2) / 2=-1+2√2

x2 = (-2-4√2) / 2=-1-2√2

b) x²+2x=3, x²+2x-3=0, D=4+12=16,√16=4

x3 = (-2+4) / 2=2/2=1

x4 = (-2-4) / 2=-6/2=-3

Otvet: x1=1+2√2, x2=1-2√2, x3=1, x4=-3