Given the solution of the equation:
[tex]x=\frac{2\pm\sqrt[]{3}}{2}[/tex]Let's check which quadratic equations has the given solution,
[tex]\begin{gathered} A)3x^2+4x-1=0 \\ \text{compare with ax}^2+bx+c=0 \\ a=3,b=4,c=-1 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-4\sqrt[]{4^2+12}}{2\cdot3} \\ =\frac{-4\pm\:2\sqrt{7}}{2\cdot\:3} \\ x_{}=\frac{-2+\sqrt{7}}{3},\: \frac{-2-\sqrt{7}}{3} \end{gathered}[/tex][tex]\begin{gathered} B)4x^2-8x+1=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4\cdot\:4\cdot\:1}}{2\cdot\:4} \\ x=\frac{-\left(-8\right)\pm\:4\sqrt{3}}{2\cdot\:4} \\ x=\frac{2\pm\sqrt{3}}{2} \end{gathered}[/tex]Hence, the correct option is B) 4x²-8x+1=0
