Hej!
[tex](2-3x)(3x+2)>3-(3x-1)^2\\\\2^2-(3x)^2>3-(9x^2-6x+1)\\\\4-9x^2>3-9x^2+6x-1\\\\-9x^2+9x^2-6x>3-1-4\\\\-6x>-2 \ \ |:(-6)\\\\x<\frac{1}{3}\\\\x\in(-\infty;\frac{1}{3})[/tex]
Pozdrawiam! - Julka
