[tex]-2x^{2}+3x+2 < 0\\\\a = -2, \ b = 3, \ c = 2\\\\\Delta = b^{2}-4ac = 3^{2}-4\cdot(-2)\cdot2 = 9+16 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\x_1 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-3+5}{2\cdot(-2)} = \frac{2}{-4} = -\frac{1}{2}\\\\x_2 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-3-5}{-4} = \frac{-8}{-4} = 2[/tex]
a < 0, to parabola zwrócona jest ramionami do dołu, wówczas:
x ∈ (-∞; -¹/₂) ∪ (2; +∞)
