Для каждого значения a решите неравенство: x^2+(2-5a)x+6a^2-3a-3<0.

\[= a^{2} - 8a + 16 =\]

\[= (a - 4)^{2} \geq 0 - при\ любом\ \]

\[значении\ переменной\ a.\]

\[x_{1} = \frac{- (2 - 5a) - (a - 4)}{2} =\]

\[= \frac{4a + 2}{2} = 2a + 1;\]

\[x_{2} = \frac{- (2 - 5a) + (a - 4)}{2} =\]

\[= \frac{6a - 6}{2} = 3a - 3.\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) < 0\]

\[x_{1} < x < x_{2}.\]

\[Сравним\ корни:\]

\[2a + 1 > 3a - 3\]

\[a < 4.\]

\[При\ a < 4:\]

\[3a - 3 < x < 2a + 1.\]

\[При\ a = 4:\]

\[x \in \varnothing.\]

\[При\ a > 4:\]

\[2a + 1 < x < 3a - 3.\]