Równanie kwadratowe ma dwa pierwiastki ujemne, gdy:
Δ > 0
x₁ + x₂ < 0
x₁ · x₂ > 0
[tex]x^{2}-2mx + m^{2}-4 = 0\\\\a = 1, \ b = -2m, \ c = m^{2}-4\\\\\Delta = b^{2}-4ac = (-2m)^{2}-4\cdot1\cdot(m^{2}-4) = 4m^{2}-4m^{2}+16 = 16 > 0[/tex]
Ze wzorów Viete'a
[tex]x_1+x_2 = \frac{-b}{2a} = -(-2m) = 2m\\\\2m < 0 \ \ \rightarrow \ \ m < 0\\\\\\x_1\cdot x_2 = \frac{c}{a} = m^{2}-4\\\\(m-2)(m+2) > 0 \ \ \rightarrow \ \ m \in (-\infty;-2) \ \cup \ (2;+\infty)[/tex]
Łącznie:
[tex]\boxed{m \in (-\infty; -2)}[/tex]
