# How to show that the quadratic equation has solutions with real part negative?

If I have a quadratic equation

$ax^2+bx+c=0$,

where a,b and c are all strictly positive.

Considering complex number solutions to the equation, how do I show that the real part of the solution is strictly negative?

My solution uses the quadratic formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Suppose $b^2-4ac \leq 0$, then the real part of the solution is $-\frac{b}{2a}$ which is strictly negative.

Suppose $b^2-4ac > 0$, then $\sqrt{b^2-4ac} < b$, so that the solution will be strictly negative.

Is there a neater way to do this without using the quadratic formula and proving by cases?

By dividing by $a$, the trinomial equation can be rewritten
$$x^2-Sx+P=0$$
Where $S$ is the sum of the roots, and $P$ is the product of the roots. This is just an instance of Vieta's formulas.
Suppose it has real roots. Here, since $a,b,c>0$, you know that $P>0$, hence the roots have same sign, and $S<0$, hence they are both negative.
What happens if the roots are complex? They are complex conjugate, so you can write them as $u\pm iv$. Their product is simply $|u+iv|^2$, and their sum is $2u$, which is negative (since $S<0$), hence $u<0$, that is, the real part of the roots is negative.