[tex]x^2+2(m-1)x+2m+1=0\\
\text{O functie de gradul 2 are solutii reale daca }\boxed{\Delta\geq 0}\\
\text{Prin urmare:}\\
4(m-1)^2-4(2m+1)\geq 0\\
4(m^2-2m+1)-8m-4\geq 0\\
4m^2-8m+4-8m-4\geq 0\\
4m^2-16m\geq 0\\
4m(m-4)\geq 0\\
\text{Dupa ce faci un tabel de semn vei obtine:}\\
m\in (-\infty,0]\cup [4,\infty)[/tex]