$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ show that $x=-c/b$ when $a=0$ OK, this one has me stumped. Given that the solution for $ax^2+bx+c =0$ $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad(*)$$ 
How would you show using $(*)$ that $x=-c/b$ when $a=0$
(Please dont use $a=0$ hence, $bx+c=0$.)

You don't. The quadratic formula only works if $a\neq 0$ (remember, if you try dividing by $0$, the universe explodes).
If $a=0$, then the equation is
$$0x^2 + bx + c = 0$$
or equivalently,
$$bx + c = 0.$$
So, solve that equation for $x$. If $b=0$, then it's always false if $c\neq 0$, and always true if $c=0$.
And if $b\neq 0$, then...

Given your comment... you could look at limiting behavior. Then
$$\begin{align*}
\lim_{a\to 0}\frac{-b\pm\sqrt{b^2-4ac}}{2a} &= \lim_{a\to 0}\frac{b^2-(b^2-4ac)}{2a\left(-b\mp\sqrt{b^2-4ac}\right)}\\
&= \lim_{a\to 0}\frac{2c}{-b\mp\sqrt{b^2-4ac}}.
\end{align*}$$
As $a\to 0$, $\sqrt{b^2-4ac}\to |b|$. If $b\gt 0$, only the minus sign in the denominator makes sense, giving you $\frac{2c}{-2b} = -\frac{c}{b}$. If $b\lt 0$, only the plus sign in the denominator makes sense, giving you again $\frac{2c}{-2b} = -\frac{c}{b}$. (The "other root" goes to $\infty$ or $-\infty$, and so is "out of sight, out of mind..."; unless $c=0$; but if $c=0$, then the original function has one value equal to $0$ and the other to $-\frac{b}{a}$; the latter goes to $\pm\infty$ as $a\to 0$, and the former has limit $0=-\frac{c}{b}$).
And if $b=0$, then of course the whole thing doesn't makes sense unless $c=0$, in which case $x$ can be anything.
