\(x^{2} - 2x \geq 3 \implies x^{2} - 2x - 3 \geq 0\)
\(x^{2} + x - 3x - 3 = (x + 1)(x - 3) \geq 0\)
\(x = -1 ; x = 3\)
Check: \(x = -1 : (-1)^{2} - 2(-1) = 1 + 2 \geq 3\) (satisfied)
\(-1 < x < 3 : 0^{2} - 2(0) = 0 \geq 3\) (not satisfied)
\(x < -1 : (-2)^{2} - 2(-2) = 4 + 4 = 8 \geq 3\) (satisfied)
\(x = 3 : 3^{2} - 2(3) = 9 - 6 = 3 \geq 3\) (satisfied)
\(x > 3 : 4^{2} - 2(4) = 16 - 8 = 8 \geq 3\) (satisfied)
\(\therefore x^{2} - 2x \geq \text{3 is satisfied in the region x} \leq \text{-1 and x} \geq 3\)