DAJE NAJ ! :D  Rozwiaz nierownosci 3-x(x+1)≤x²+4x x²≥5

1) [latex]3-x(x+1) leq x^2+4x \ 3-x^2-x leq x^2+4x\ -2x^2-5x+3 leq 0\ Delta=1 sqrtDelta=1\ x_1=-1 x_2=- frac{3}{2} \ xin(-infty, - frac{3}{2} extgreater cup extless -1,+ infty)\[/latex] 2) [latex]x^2 geq 5\ x^2-5 geq 0\ (x- sqrt5)(x+ sqrt5) geq0\ x in(-infty,-sqrt5)cup(sqrt5,+infty) [/latex]