Try it, please~

if we transpose $3x - 1$ we get, $2x^2 - 3x + 1$ $= 2x^2 - 2x - 1x + 1$ $= 2x(x - 1) + 1 (x - 1)$ $= (x - 1) (2x - 1)$ therefore we get two solutions $x = 1$ and $x = 1/2$

Also, please type your latex in in brackets form so that it becomes, you know that.....

$2x^2=3x-1$ can be written as

$2x^2-3x+1=0$

Using the quadratic formula to solve for x, we get

$x=1$

$x=\dfrac{1}{2}$