usor/.............
2[tex]4 x^{2} + y^{2} +4x+6y+10=0 => (2x)^{2} + y^{2} + 2*2x*1+2*3*y+9+
[/tex]+1=[tex](2x+1)^{2} +(Y+3)^{2} =0[/tex]
dar [tex] (2x+1)^{2} \geq 0[/tex], oricare ar fi x care apartine R
si [tex] (y+3)^{2} \geq 0[/tex], oricare ar fi y care apartine R .
de unde rezulta ca [tex] (2x+1)^{2} =o=> 2x+1=0=> x= \frac{-1}{2} [/tex]
[tex] (y+3)^{2} =0=> y+3=0=> y=-3[/tex]
