Rozwiąż równanie (2-x)^2 - 5 >/ (większe lub równe) (x-pierw.z z 3)^2

[latex](2-x)^2 - 5 geq (x- sqrt{3})^2\\4-4x+x^2-5geq x^2-2sqrt{3}x+3\\ -4x+x^2- x^2+2sqrt{3}x geq3-4+5 \\ -4x +2sqrt{3}x geq4 /*(-1) \\ 4x -2sqrt{3}x leq -4\\x(4-2sqrt{3})leq 6 /:(4-2sqrt{3})[/latex] [latex] x leq frac{-4}{4-2sqrt{3}}*frac{4+2sqrt{3}}{4+2sqrt{3}}\\xleq frac{-4( 4+2sqrt{3})}{4^2-(2sqrt{3})^2}\\xleq frac{ 4( -4-sqrt{3})}{16-12} \\xleq frac{4( -4-2 sqrt{3})}{4} \\xleq -4- 2sqrt{3} \\xin extless (-4-2sqrt{3});-infty )[/latex]