0 \\ D=4+12=16 \\ x_1=\frac{-2-4}{6}=-1 \\ x_2=\frac{-2+4}{6}=\frac{1}{3} \\ 3(x+1)(x-\frac{1}{3}) > 0 " alt="3x^2+2x-1 > 0 \\ D=4+12=16 \\ x_1=\frac{-2-4}{6}=-1 \\ x_2=\frac{-2+4}{6}=\frac{1}{3} \\ 3(x+1)(x-\frac{1}{3}) > 0 " align="absmiddle" class="latex-formula">
---(+)--(○-1)--(-)----(○1/3)----(+)--->
