Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$ 
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$

Hint: Start by dividing the whole equation by $a$
At first I have tried solving the equation without using the hint provided in my exercise and directly applying completing square, I get $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-c}{a}}$. So if I am to use the hint, I obtain the appropriate answer. But I wonder if I am asked the same question in my exam where the hint will not be provided then how am I  supposed to answer.
I would like to know how one should approach this kind of question and how do I realise when to divide the whole equation with in this case $a$ or is there any other ways so that I can avoid dividing the whole equation by $a$.Thanks in advance for any help you are able to provide!
EDIT: Here's my steps. Please see where have I done wrong.
\begin{align}
ax^2+2bx+c&=0 \\
a\left[\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a^2}\right] + c&=0 \\
a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a} + c&=0 \\
\left(x+\frac{b}{a}\right)^2&=\left(\frac{b^2}{a}-c\right)\left(\frac{1}{a}\right)\\
\left(x+\frac{b}{a}\right)^2&=\frac{a(b^2-c)}{a^2}\\
\left(x+\frac{b}{a}\right)^2&=\frac{b^2-c}{a} \\
x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-c}{a}} \\
\implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-c}{a}}\\
\end{align}

HINT: Complete the square:
$$ax^2+2bx+c=0$$
$$a\left(x^2+\frac{2b}{a}x\right)+c=0$$
Try the rest from here.
