Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have 
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots: 
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-write the original quadratic as: 
$$(x-1)\left(x-\tfrac{1}{2}\right)$$
However, if you actually multiply this out then you get: 
$x^2-\frac{3}{2}x + \frac{1}{2}$ which is not the same as  $2x^2-3x+1$. 
So why the discrepancy? Aren't you supposed to get the original equation? 

Since you are applying quadratic formula to factorize, I am assuming that you made a typo in the degree of x in the first term. It should be $2x^2$ and not $2x^3$ 
Also,
$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is the correct factorization of the trinomial.
$f(x)=0$ does have roots $x_1$ and $x_2$, but the multiplicand '$a$' $=2$ is missing in your reconstruction of the trinomial as a product of its factors.
