Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]x(x-6) > 5(1-2x)\\\\x^{2}-6x>5-10x\\\\x^{2}-6x+10x-5 > 0\\\\x^{2}+4x-5> 0\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot1\cdot(-5) = 16+20 = 36\\\\\sqrt{\Delta} = \sqrt{36} = 6\\\\x_1 = \frac{-4-6}{2} =-\frac{10}{2} = -5\\\\x_2 = \frac{-4+6}{2}=\frac{2}{2} = 1\\\\a > 0, to \ ramiona \ paraboli \ skierowane \ do \ gory, \ wowczas\\\\x \in(0;-5) \ \cup \ (1;+\infty)[/tex]
