Equation has two distinct solutions We have $\displaystyle{ a,b,c }$ real numbers such that $\displaystyle{ a^2+b^2+c^2>0 }$. Which condition must hold so that the equation $\displaystyle{ ax^2+bx+c=0 }$ has two different solutions?
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I have done the following :
\begin{align*}&ax^2+bx+c=0 \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0 \text{ if } a\neq 0 \\ &a=0 \Rightarrow ax^2+bx+c=0 \Rightarrow bx+c=0 \Rightarrow x=-\frac{c}{b}\ \text{ In this case we have only one solution, so it must be } a\neq 0\end{align*}
We have that \begin{equation*}x^2+\frac{b}{a}x+\frac{c}{a}=0 \Rightarrow x=-\frac{b}{2a}\pm \sqrt{\left (\frac{b}{2a}\right )^2-\frac{c}{a}}\end{equation*}
We have two different solution if $\frac{b^2}{4a^2}\neq \frac{c}{a}$, i.e. if $b^2\neq 4ac$.
Is the condition correct?

for any degree two polynomial in order for it to have real roots we must have :
$$\Delta = b^2-4ac > 0$$
and if $\Delta \neq 0$ it always has two distinct solutions real or complex so that I believe suffices.
