Помогите с алгеброой пж

2x^2 - 5x + 2 < 0
___ 2x^2 - 5x + 2 = 0
x(1,2) = [5 + - V(5^2 - 4*2*2)]/2*2 = (5 + - 3)/4
x1 = (5-3)/4 = 1/2
x2 = (5+3)/4 = 2
=>
2*(x - x1)(x - x2) < 0 или
(x - 1/2)(x - 2) < 0
x > 1/2 и x < 2
1/2 < x < 2

3x - x^2 >= 0
x * (3 - x) >= 0
x > 0 => (3-x) > 0 или x >= 0 и x =< 3
0 =< x =< 3

x^2 > 121
x^2 - 11^2 > 0x
(x + 11)(x - 11) > 0
или
{ (x+11) < 0 ---> x < - 11
{ (x-11) < 0 ---> x < 11
или
{ (x+11) > 0 ---> x > - 11
{ (x-11) > 0 ---> x > 11
x < - 11 или x > 11