Solve the equation $2x^{2}-5x+3=0$ by the method of completing the square.

(N/A) The given equation is $2x^{2}-5x+3=0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2}-\frac{5}{2}x+\frac{3}{2}=0$
Now,complete the square by adding and subtracting the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{5}{2})^{2} = (\frac{5}{4})^{2} = \frac{25}{16}$:
$x^{2}-\frac{5}{2}x + \frac{25}{16} - \frac{25}{16} + \frac{3}{2} = 0$
$(x-\frac{5}{4})^{2} - \frac{25}{16} + \frac{24}{16} = 0$
$(x-\frac{5}{4})^{2} - \frac{1}{16} = 0$
$(x-\frac{5}{4})^{2} = \frac{1}{16}$
Taking the square root on both sides:
$x-\frac{5}{4} = \pm \frac{1}{4}$
Case $1$: $x-\frac{5}{4} = \frac{1}{4} \implies x = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$
Case $2$: $x-\frac{5}{4} = -\frac{1}{4} \implies x = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1$
Thus,the solutions are $x = \frac{3}{2}$ and $x = 1$.