a]
[tex]2x^2+16x+30 > 0 \ \ |:2\\\\x^2+8x+15 > 0\\\\a=1, \ b=8, \ c=15\\\\\Delta=b^2-4ac=8^2-4\cdot1\cdot15=64-60=4\\\\\sqrt{\Delta}=\sqrt4=2\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\Rightarrow\frac{-8-2}{2\cdot1}=\frac{-10}{2}=-5\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\Rightarrow\frac{-8+2}{2\cdot1}=\frac{-6}{2}=-3\\\\\huge\boxed{x\in(-\infty;-5)\cup(-3;+\infty)}[/tex]
a > 0, więc ramiona paraboli są skierowane do góry
b]
[tex]\frac{x+3}{x-4}=\frac{x-5}{x+2}\\\\(x+3)(x+2)=(x-4)(x-5)\\\\x^2+2x+3x+6=x^2-5x-4x+20\\\\x^2+5x+6=x^2-9x+20\\\\x^2+5x-x^2+9x=20-6\\\\14x=14 \ \ |:14\\\\\huge\boxed{x=1}[/tex]
c]
[tex]\frac{x+2}{4}=\frac{3x+5}{3}-\frac{4x+5}{6} \ \ |\cdot12\\\\3(x+2)=4(3x+5)-2(4x+5)\\\\3x+6=12x+20-8x-10\\\\3x+6=4x+10\\\\4x-3x=6-10\\\\\huge\boxed{x=-4}[/tex]
