He have the following:
[tex]4x^2+12x>-x-3[/tex]solving:
[tex]\begin{gathered} 4x^2+12x+x+3>-x-3+x+3 \\ 4x^2+13x+3>0 \\ \text{factoring} \\ (4x+1)(x+3)>0 \\ (4x+1)>0\rightarrow4x+1-1>-1\rightarrow\frac{4x}{4}>-\frac{1}{4}\rightarrow x>-\frac{1}{4} \\ (x+3)<0\rightarrow x+3-3<0-3\rightarrow x<-3 \\ \text{the solution is} \\ (-\infty,-3)U(-\frac{1}{4},\infty) \end{gathered}[/tex]