<s>[INST] 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? 
 [/INST] The quadratic formula gives the solution to the equation
$$ax^2+bx+c=0.$$
But the formula is derived by completing the square, and to do so the coefficient of $x^2$ must be 1. Therefore, the first step is to divide by $a$ (it is assumed without loss of generality that $a>0$). This gives
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$$
Now notice that 
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=\left(x-\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\right)\left(x-\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\right).$$
Long story short, the quadratic formula gives the solutions of $ax^2+bx+c=0$ by finding the solutions to the $equivalent$ equation $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$
</s>