\(2x=\sqrt{3x-1}\implies (2x)^2=(\sqrt{3x-1})^2\implies 4x^2=3x-1\implies 4x^2-3x+1=0\)
You can use the quadratic formula to solve for \(x\), but first check the discriminant:
\(\Delta(ax^2+bx+c)=b^2-4ac\implies \Delta(4x^2-3x+1)=-27\)
Since the discriminant is negative, this quadratic equation has two complex solutions, so there is no real solution to the original equation.
