When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:

$$ax^2+bc+c=0$$

I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}*\frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
 A: "The common denominator will be the product of the denominators so $4a^3$" -- this is wrong.
The common denominator is $4a^2$, not $4a^3$. You multiply denominators only when they don't have a common factor; here they do.
A: Note that
$$
\frac{c}{a}=\frac{4ac}{4a^2}
$$
whence
$$
\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}
$$
In your approach
$$
 \frac{b^2a-4a^2c}{4a^3}=\frac{b^2-4ac}{4a^2}.
$$
A: Here is an alternate approach which avoids working with fractions
\begin{eqnarray}
ax^2+bx+c&=&0\\\
&&\\\
\text{subtract }c&&\\\
ax^2+bx&=&-c\\\
&&\\\
\text{multiply by }4a&&\\\
4a^2x^2+4abx&=&-4ac\\\
&&\\\
\text{add }b^2&&\\\
4a^2x^2+4abx+b^2&=&b^2-4ac\\\
&&\\\
\text{factor the left side}&&\\\
(2ax+b)^2&=&b^2-4ac\\\
&&\\\
\text{take square root of both sides}&&\\\
2ax+b&=&\pm\sqrt{b^2-4ac}\\\
&&\\\
\text{solve for }x&&\\\
x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{eqnarray}
A: In your last step $\frac{b^2a - 4a^2c}{4a^3}$ cancel $a$ from numerator and denominator, to get $ \frac{b^2 - 4ac}{4a^2}$
or take the common denominator as $4a^2$ in the beginning itself.
