X^2-2x-3>0 решите пож очень нужно

0 \\ x^{2} -2x-3=0 \\ D=4+12=16 \\ \sqrt{D} =4 \\ x_{1} = \frac{2+4}{2} = \frac{6}{2} =3 \\ x_{2} = \frac{2-4}{2} = \frac{-2}{2} =-1" alt=" x^{2} -2x-3>0 \\ x^{2} -2x-3=0 \\ D=4+12=16 \\ \sqrt{D} =4 \\ x_{1} = \frac{2+4}{2} = \frac{6}{2} =3 \\ x_{2} = \frac{2-4}{2} = \frac{-2}{2} =-1" align="absmiddle" class="latex-formula">

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x∈ $(- \infty ;-1)U(3;+ \infty )$
Ответ: x∈ $(- \infty ;-1)U(3;+ \infty )$