3·t² -a·t +4=0
Δ= a² -3·4·4= a² -48
daca Δ >0 atunci ecuatia de gradul II are exact doua elemente , adica doua solutii
a² -48 >0
( a- 4√3) ( a+4√3) >0
a -∞ -4√3 4√3 ∞
---------------------------------------------------------------------------------------------------
a²-48 + + 0 - - - 0 + +
a∈ ( -∞ , -4√3) U ( 4√3 , +∞)