Odpowiedź :
Odpowiedź:
-3x² + x + 1 ≥ 0
Δ = 1 - 4 * (-3) * 1 = 13
[tex]\sqrt{\Delta} = \sqrt{13}[/tex]
[tex]x = \frac{4 - \sqrt{13}}{2 * (-3)}[/tex] v [tex]x = \frac{4+\sqrt{13}}{2*(-3)}[/tex]
[tex]x = -\frac{4-\sqrt{13}}{6}[/tex] v [tex]x = -\frac{4+\sqrt{13}}{6}[/tex]
[tex]x \in < -\frac{4+\sqrt{13}}{6}, -\frac{4-\sqrt{13}}{6} >[/tex]