Question

-3x^2+14x-8≥0
ПОМОГИТЕ Я ЩПС УМРУ ​

Answer 1

-3x^2+14x-8≥0
-3х^2+12х+2х-8 ≥ 0
-3х • (х-4)+2(х-4) ≥ 0
-(х-4)(3х-2) ≥ 0
(х-4) (3х-2) ≤ 0
{х-4 ≤ 0 ; х ≤ 4
{3х-2 ≤ 0; х ≤ 2/3
х ∈ [2/3;4]