It is well known fact that given $f(x)=ax^2+bx+c, \ a \neq 0$ and $\Delta=b^2-4ac>0$, then $f(x)$ has
two distinct roots.I assume we are in $\mathbb{R}$ at any stage of the problem.
My Attempt
To prove this I considered that
$f(x)=a \left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right) \left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)$
upon applying Gaussian completion of squares.
Then I said: suppose the roots are not distinct, then $-\sqrt{\Delta}=\sqrt{\Delta}$. This will mean $\Delta=0$. Thus we remain with choice $\Delta \neq 0$.
From here, I don't know how to get to next stage.
Best Answer
Starting from
$$ax^2+bx+c$$ if $a\ne 0$ we have $$ax^2+bx=-c \iff x^2+\frac bax\color{red}{+\frac{b^2}{4a^2}}=\color{red}{\frac{b^2}{4a^2}}-\frac ca$$ $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\equiv \frac{\Delta}{4a^2}$$ $$\left(x+\frac{b}{2a}\right)^2= \frac{\Delta}{4a^2} \tag 1$$ If $\Delta <0$, there are no real solutions since the left-hand side is non-negative and $4a^2>0$; if $\Delta=0$, we have $$\left(x+\frac{b}{2a}\right)^2=0\implies x=-\frac{b}{2a}$$ The solutions are $x=-\frac{b}{2a}$ (multiplicity 2). If $\Delta >0$, eq. $(1)$ becomes
$$\left(x+\frac{b}{2a}\right)^2- \left(\frac{\sqrt{\Delta}}{2a}\right)^2 =0,$$ i.e.,
$$x_1+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}=0 \quad \vee \quad x_2+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}=0$$ which are the distinct solutions of an equation of degree 2.