[tex](2x+3)^{2}-9 > 8-(3-x)^{2}\\\\4x^{2}+12x+9-9 > 8-(9-6x+x^{2})\\\\4x^{2}+12x > 8-9+6x-x^{2}\\\\4x^{2}+x^{2}+12x-6x+1 > 0\\\\5x^{2}+6x+1 > 0\\\\a = 5, \ b = 6, \ c = 1\\\\\Delta = b^{2}-4ac = 6^{2}-4\cdot5\cdot1 = 36-20 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} =\frac{-6-4}{2\cdot5} = \frac{-10}{10} = -1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} =\frac{-6+4}{10} = \frac{-2}{10} = -\frac{1}{5}[/tex]
a > 0, to parabola zwrócona jest ramionami do góry, wówczas:
[tex]\boxed{x \in (-\infty; -1) \ \cup \ (-\frac{1}{5}; +\infty)}[/tex]
