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?