$3x^2+3x-1=0\\ \Leftrightarrow \left(\sqrt{3}x\right)^2+\dfrac{3}{2}x+\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{7}{4}=0\\ \Leftrightarrow \sqrt{3}x\left(\sqrt{3}x+\dfrac{\sqrt{3}}{2}\right)+\dfrac{\sqrt{3}}{2}\left(\sqrt{3}x+\dfrac{\sqrt{3}}{2}\right)-\dfrac{7}{4}=0\\ \Leftrightarrow \left(\sqrt{3}x+\dfrac{\sqrt{3}}{2}\right)^2=\dfrac{7}{4}\\ \Leftrightarrow \left|\sqrt{3}x+\dfrac{\sqrt{3}}{2}\right|=\dfrac{\sqrt{7}}{2}(*)\\ \circledast \sqrt{3}x+\dfrac{\sqrt{3}}{2}\ge 0 \Leftrightarrow x\ge-\dfrac{1}{2}\\(*) \Leftrightarrow \sqrt{3}x+\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{7}}{2}\\ \Leftrightarrow \sqrt{3}x=\dfrac{\sqrt{7}-\sqrt{3}}{2}\\ \Leftrightarrow x=\dfrac{\sqrt{7}-\sqrt{3}}{2\sqrt{3}}\\ \Leftrightarrow x=\dfrac{\sqrt{21}-3}{6}(TM)\\ \circledast \sqrt{3}x+\dfrac{\sqrt{3}}{2}<0 \Leftrightarrow x<-\dfrac{1}{2}\\ (*) \Leftrightarrow-\sqrt{3}x-\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{7}}{2}\\ \Leftrightarrow -\sqrt{3}x=\dfrac{\sqrt{7}+\sqrt{3}}{2}\\ \Leftrightarrow x=-\dfrac{\sqrt{7}+\sqrt{3}}{2\sqrt{3}}\\ \Leftrightarrow x=-\dfrac{\sqrt{21}+3}{6}(TM)$
