[x]=partea intreaga a numarului x . fie [x]=k∈[-4 ,1] k=numar intreg
3k²+4k-4=0
Δ=16-4*3*(-4)=16+48=64>0
k1=(-4-√64)/2*3=(-4-8)/6= -2
k2=(-4+8)/6=2/3∉Z
Deci
[x]=-2 +> x∈[-2 , -1) ecuatia are o solutie
b) este aceiasi ecuatie dar in loc de partea intreaga ai partea fractionara
{x}=y y∈[0,1)
3y²+4y-4=0
y1=-2 nu se accepta ramane
y=2/3 =0,(6)
{y}=0,666....
