Odpowiedź:
Szczegółowe wyjaśnienie:
-x²+5x-6>0
Δ=b²-4ac
a= -1,b=5,c=-6
Δ=5²- 4*(-1)*(-6)=25-24=1
√Δ=1
x1=-b-√Δ/2a
x1= -5-1/2*(-1)=-6/-2=3
x2=-b+√Δ/2a
x= -5+1/2*(-1)= -4/-2=2
x∈(2,3) -> ramiona paraboli skierowane w dół, bo a< 0
[tex]-x^{2}+5x-6 > 0\\\\a = -1, \ b = 5, \ c = -6\\\\\Delta = b^{2}-4ac = 5^{2}-4\cdot(-1)\cdot(-6) = 25-24 = 1\\\\\sqrt{\Delta} = \sqrt{1} = 1\\\\x_1 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-5+1}{2\cdot(-1)} = \frac{-4}{-2} = 2\\\\x_2 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-5-1}{-2} = \frac{-6}{-2}=3\\\\a < 0, \ to \ ramioa \ wykresu \ paraboli \ skierowane \ do \ dolu, \ wowczas:\\\\\boxed{x\in (2;3)}[/tex]