[tex]3x^2+14x-5>0\\3x^2-x+15x-5>0\\x(3x-1)+5(3x-1)>0\\(x+5)(3x-1)>0\\x\in(-\infty,-5)\cup\left(\dfrac{1}{3},\infty\right)[/tex]
[tex]-6x^2=-x-1\\6x^2-x-1=0\\6x^2-3x+2x-1=0\\3x(2x-1)+1(2x-1)=0\\(3x+1)(2x-1)=0\\x=-\dfrac{1}{3} \vee x=\dfrac{1}{2}[/tex]
