Solve $7x^2+3x=0$ by completing the square? I am to solve for x: $7x^2+3x=0$
I'm aware that there are multiple approaches to solving a quadratic. In this case, since there is no constant term I decided to go the completing the square route. I know from my textbooks answer that the solutions are $x=0$ and $x=-\frac{3}{7}$.
Here is how far I got:
$7x^2+3x=0$ # want to have leading coefficient 1 not 7
$x^2 + \frac{3}{7}x=0$
take 1/2 of the linear coefficient and then square it:
$\frac{1}{2}*\frac{3}{7}=\frac{3}{14}$
Then square it:
$(\frac{3}{14})^2$ = $\frac{9}{196}$
Add this term to both sides of my equation:
$x^2 + \frac{3}{7}x + \frac{9}{196}=\frac{9}{196}$
This is where I get stuck. Apparently I should be able to factor as a perfect square the left hand side of the equation. Perhaps because I'm working with fractions I cannot see how to do that? How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?

$$7x^2+3x=0 \iff 7(x^2+(3/7)x)=0\iff$$ $$\iff 7 (\;(x+(1/2)(3/7)\,)^2 -(\,(1/2)(3/7)\,)^2\;)=0\iff $$ $$\iff (x+(1/2)(3/7)\,)^2-(\,(1/2)(3/7)\,)^2=0 \iff$$ $$\iff  (x+(1/2)(3/7)\,)^2=(\,(1/2)(3/7)\,)^2\iff$$ $$\iff  x+(1/2)(3/7)=\pm (1/2)(3/7)\iff$$ $$\iff (\;(x+(1/2)(3/7)=(1/2)(3/7)\; \lor \;x+(1/2)(3/7)=-(1/2)(3/7)\;) \iff$$ $$\iff   (\,x=0\,\lor \,x=-3/7\,).$$
Of course if $A\ne 0$ then $Ax^2+Bx=0 \iff A(x)(x+B/A)=0 \iff (x=0 \lor x+B/A=0)\iff (x=0\lor x=-B/A).$
