Odpowiedź :
[tex]3x+3-x^{2} -x+5>0\\-x^{2}+2x+8>0\\delta = 4 - 4*(-1)*8 = 36\\\\\sqrt{delta} = 6\\x_{1} = \frac{-2-6}{-2}=\frac{-8}{-2}=4 \\\\x_{2} =\frac{-2+6}{-2}=\frac{4}{-2}=-2\\\\\\Odpowiedz: x ∈ (-2, 4)[/tex]
[tex]3x+3-x^{2} -x+5>0\\-x^{2}+2x+8>0\\delta = 4 - 4*(-1)*8 = 36\\\\\sqrt{delta} = 6\\x_{1} = \frac{-2-6}{-2}=\frac{-8}{-2}=4 \\\\x_{2} =\frac{-2+6}{-2}=\frac{4}{-2}=-2\\\\\\Odpowiedz: x ∈ (-2, 4)[/tex]