Nu inteleg aici de unde o aparut a₁=1 si a₂= 13/5... Am nevoie de ajutor si ceva explicatii.

formula funcţiei de gradul 2 ar fi f(x)=ax^2-bx+x=0
in cazul nostru a=a+1
b=3(a-1)=3a-3
c=a

delta=b^2-4ac=(3a-3)^2-4×(a+1)×(a-1)=9a^2-18a+9-4×(a^2-1)=9a^2-18a+9-4a^2+4=5a^2-18a+13
delta=b^2-4ac=18^2-4×5×13=324-260=64

a1=(-b+√delta)/2a=(18+√64)/2×5=(18+8)/10=26/10=13/5

a2=(-b-√delta)/2a=(18-√64)/2×5=(18-8)/10=10/10=1