Solve (D^3-D^2-D-2)y=0

(^3-^2+2)D=0

D^2+D^2+2D=0

2D^2+2D=0

a = 2; b = 2; c = 0;

Δ = b2-4ac

Δ = 22-4·2·0

Δ = 4

D1=−b−Δ√2aD2=−b+Δ√2a

Δ−−√=4√=2

D1=−b−Δ√2a=−(2)−22∗2=−44=−1

D2=−b+Δ√2a=−(2)+22∗2=04=0

loloo

step-by-step explanation:

lololool

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2