3х²-10х+7>4
3х²-10х+7-4>0
3x²-10x+3>0 3x²-10x+3 = 0
х∈(-∞;1/3)∪(3;+∞) D = 100-4*3*3 = 64
x = (10+8)/6 = 3
x= (10-8)/6 = 1/3
одз:
Зх²-10х+7>0 3х²-10х+7=0
x∈(-∞;1)∪(2 1/3;+∞) D = 100-4*3*7 = 16
x = (10-4)/6 = 1
x = (10+4)/6 = 2 1/3
Ответ: х∈(-∞;1/3)∪(3;+∞)