Explicație pas cu pas:
(x²+x+1)(x²+x+2)=12
Notam x²+x=t
(t+1)(t+2)=12=>
t²+3t+2=12=>t²+3t-10=0 => t²+5t-2t-10=0=>
t(t+5)-2(t+5)=0=>
(t+5)(t-2)=0=>t1=2 sau t2=-5
x²+x=2=>x²+x-2=0=>x²+2x-x-2=0=>x(x+2)-1(x+2)=0=>
(x-1)(x+2)=0=>x=1 sau x=2
x²+x=-5=>x²+x+5=0 Fals pt ca avem /\<0 deci nu are solutii reale
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