Найдите целые системы решений неравенств: 2x^2-3x-5<=0; -x^2+8x-12<0.

Ответ:

\[1)\ 2x^{2} - 3x - 5 =\]

\[= 2 \cdot (x + 1)(x - 2,5)\]

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{3 + 7}{4} = \frac{10}{4} = 2,5;\]

\[x_{2} = \frac{3 - 7}{4} = - \frac{4}{4} = - 1.\]

\[2 \cdot (x + 1)(x - 2,5) \leq 0\]

\[- 1 \leq x \leq 2,5.\]

\[2)\ x^{2} - 8x + 12 =\]

\[= (x - 2)(x - 6)\]

\[D_{1} = 16 - 12 = 4\]

\[x_{1} = 4 + 2 = 6;\ \ x_{2} = 4 - 2 = 2.\]

\[- (x - 2)(x - 6) < 0\]

\[(x - 2)(x - 6) > 0\]

\[x < 2;\ \ x > 6.\]

\[Ответ:\ - 1 \leq x < 2.\]