Question

How to solve for X for 2X= 3X-1 when 3X-1 is under a square root

Answer 1

$2x=\sqrt{3x-1}\implies (2x)^2=(\sqrt{3x-1})^2\implies 4x^2=3x-1\implies 4x^2-3x+1=0$

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.