[tex](x-5)^2>2(5-x)\\\\x^2-2\cdot x\cdot5+5^2>10-2x\\\\x^2-10x+25>10-2x\\\\x^2-10x+25-10+2x>0\\\\x^2-8x+15>0\\\\a=1, \ b=-8, \ c=15\\\\\Delta=(-8)^2-4\cdot1\cdot15=64-60=4\\\\\sqrt{\Delta}=\sqrt4=2\\\\x_1=\frac{-(-8)-2}{2\cdot1}=\frac{6}{2}=3\\\\x_2=\frac{-(-8)+2}{2\cdot1}=\frac{10}{2}=5\\\\x\in(-\infty;3)\cup(5;+\infty)[/tex]
a > 0, ramiona paraboli są skierowane do góry, kółeczka niezamalowane
[tex](x - 5)^{2} > 2(5-x)\\\\(x-5)^{2} -2(5-x) > 0\\\\(x-5)^{2} +2(x-5) > 0\\\\(x-5)(x-5+2) =0\\\\(x-5)(x-3) = 0\\\\x-5 = 0 \ \vee \ x-3 =0\\\\x = 5 \ \vee \ x = 3\\\\a > 0, \ to \ ramiona \ paraboli \ skierowane \ do \ gory\\\\x \in (-\infty; 3) \ \cup \ (5;+\infty)[/tex]