Answer:[tex]\large\boxed{x=\dfrac{-4-2\sqrt7}{3}\ or\ x=\dfrac{-4+2\sqrt7}{3}}[/tex]Step-by-step explanation:[tex]\text{The quadratic formula for a quadratic equation}\ ax^2+bx+c=0\\\\\text{determinant:}\ \Delta=b^2-4ac\\\\\text{If}\ \Delta<0,\ \text{then the equation has no solution.}\\\\\text{If}\ \Delta=0,\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}.\\\\\text{If}\ \Delta>0,\ \text{then the equation has two different solutions}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}.[/tex][tex]\text{We have:}\\\\3x^2+8x-4=0\\\\a=3,\ b=8,\ c=-4\\\\\Delta=8^2-4(3)(-4)=64+48=112>0\\\sqrt\Delta=\sqrt{112}=\sqrt{(16)(7)}=\sqrt{16}\cdot\sqrt7=4\sqrt7\\\\x=\dfrac{-8\pm4\sqrt7}{2(3)}=\dfrac{-8\pm4\sqrt7}{6}=\dfrac{2(-4\pm2\sqrt7)}{6}=\dfrac{-4\pm2\sqrt7}{3}[/tex]