Răspuns :
determinantul ecuatiei Δ=1²-4*3*(-2)=1+24=25
x1=(-1-√25)/2*3=(-1-5)/6=-6/6= -1∈Z
x2=(-1+√25)/2*3=(-1+5)/6=4/6=2/3∉Z
Deci p(x)=fals
3x²+x-2=0
Δ=b²-4ac=1-4·3·(-2)=1-(-24)=25 => x₁=-1-5 /6 => x₁=-6/6=-1 sau x₂=-1+5 /6 => x₂=4/6=2/3 => 1∈Z, iar 2/3∈Q\Z ,asadar propozitia este falsa.