Найдём дискриминант:
\(D = {\left( {a-5} \right)^2}-4 \cdot \left( {-2{a^2} + 2a + 4} \right) = {a^2}-10a + 25 + 8{a^2}-8a-16 = 9{a^2}-18a + 9 = 9{\left( {a-1} \right)^2}.\)
Если \(D = 9{\left( {a-1} \right)^2} = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a = 1,\) то неравенство примет вид:
\({x^2}-4x + 4 < 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {x-2} \right)^2} < 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .\)
Если \(a \ne 1\), то: \({x_1} = \dfrac{{-a + 5 + 3a-3}}{2} = a + 1;\,\,\,\,\,\,\,{x_1} = \dfrac{{-a + 5-3a + 3}}{2} = -2a + 4.\)
Если \({x_1} > {x_2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a + 1 > -2a + 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a > 1,\) то \(x\, \in \,\left( {-2a + 4;a + 1} \right).\)
Если \({x_1} < {x_2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a + 1 < -2a + 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a < 1,\) то \(x\, \in \,\left( {a + 1;-2a + 4} \right).\)
ОТВЕТ: \(x \in \left( {a + 1;\,4-2a} \right)\) при \(a \in \left( {-\infty ;\,1} \right);\) нет решений при \(a = 1;\) \(x \in \left( {4-2a;\,a + 1} \right)\) при \(a \in \left( {1;\,\infty } \right).\)