a) x²-2x+1=0 <=> (x-1)² =0 <=> x-1=0 => x = 1
b)x²-4x-12=0 => x²-6x+2x-12=0 => x(x-6)+2(x-6)=0 => (x+2)(x-6)=0 => x∈{-2;6}
c)2x²-6x=0 => 2(x²-3x)=0 => x²-3x=0 =>x(x-3)=0 => x∈{0;3}
d)x²+x+2=0 <=> a=1 b=1 c=2
Δ=b²-4ac=1-4·2·1=-7<0 => NU EXISTA SOLUTII REALE
Tot ce am scris e 100% corect!
