[tex](m+3)x^2-2(m-1)x+m-2=0\\
Pentru\ a\ avea\ solutii\ reale\ \Delta\ trebuie\ sa\ fie\ \geq 0.
Asadar:\\
{[2(m-1)]^2-4(m+3)(m-2) \geq 0}\\
4(m^2-2m+1)-4(m^2+3m-2m-6)\geq 0\\
4m^2-8m+4-4(m^2+m-6)\geq0\\
4m^2-8m+4-4m^2-4m+24 \geq 0 \\
-12m+28\geq 0\\
12m \leq 28 \\
m \leq \frac{7}{3}\\
S:m\in (-\infty,\frac{7}{3}][/tex]
