O ecuatie de gradul doi nu are solutii reale daca Δ < 0.
Δ = b^{2}-4ac
Iar in aceasta ecuatie:
mx^{2}-2(m+1)x+m-5=0
b = -2(m+1)
a = m
c = m-5
Deci, ecuatia nu are solutii reale daca
(-2(m+1))^{2}-4m(m-5)\ \textless \ 0
4(m+1)^{2}-4m(m-5) \ \textless \ 0
4m^{2}+8m+4-4m^{2}+20m\ \textless \ 0
28m+4\ \textless \ 0
28m\ \textless \ -4
m\ \textless \ -\frac{4}{28}
m\ \textless \ -\frac{1}{7}
sper ca te-am ajutat! intreaba-ti parintii daca e corect, mai verifica-te cu ei :] paa!
