Cześć!
a)
[tex](2x-4)^2<4\\\\(2x)^2-2\cdot2x\cdot4+4^2-4<0\\\\4x^2-16x+16-4<0\\\\4x^2-16x+12<0 \ \ /:4\\\\x^2-4x+3<0\\\\a=1, \ b=-4, \ c=3\\\\\Delta=(-4)^2-4\cdot1\cdot3=16-12=4\\\\\sqrt{\Delta}=\sqrt4=2\\\\x_1=\frac{-(-4)-2}{2\cdot1}=\frac{4-2}{2}=\frac{2}{2}=1\\\\x_2=\frac{-(-4)+2}{2\cdot1}=\frac{4+2}{2}=\frac{6}{2}=3\\\\\boxed{x\in(1;3)}[/tex]
a > 0, ramiona paraboli są skierowane do góry.
b)
[tex]3(2x+3)(x+2)>0 \ \ /:3\\\\(2x+3)(x+2)>0\\\\2x+3=0 \ \ \ \vee \ \ \ x+2=0\\\\2x=-3 \ \ \ \vee \ \ \ x=-2\\\\x=-1,5 \ \ \ \vee \ \ \ x=-2\\\\\boxed{x\in(-\infty;-2)\cup(-1,5;+\infty)}[/tex]
a > 0, ramiona paraboli są skierowane do góry.
