x^2-4x-12>0 \\x^2-4x-12=0 \\ \\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{4-\sqrt{ (-4)^2-4\cdot 1\cdot (-12) }}{2 }=\frac{4-\sqrt{16+48 }}{2}=\\\\=\frac{4-\sqrt{64 }}{2}= \frac{4-8}{2}=\frac{-4}{2}=-2\\\\ x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{4+\sqrt{ (-4)^2-4\cdot 1\cdot (-12) }}{2 }= \frac{4+8}{2}=\frac{12}{2}=6\\\\ Answer : \ \ x < -2\ \ \ or \ \ x > 6
