\(4x^2-2\sqrt{3}x+\sqrt{3}-1=0\)
\(\Delta'=3-4\left(\sqrt{3}-1\right)=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{\sqrt{3}+2-\sqrt{3}}{4}=\frac{1}{2}\\x_2=\frac{\sqrt{3}-2+\sqrt{3}}{4}=\frac{\sqrt{3}-1}{2}\end{matrix}\right.\)