Why is the root of a square restrictive in my calculation?

Question

I have a parabola $f(x) = ax^2 + bx + c$, where $a<0$, and I want to find the domain where $f$ is greater than a secant defined by taking a point $w$ on the domain, and drawing the secant from $(w-h, f(w-h))$ to $(w+h, f(w+h))$. To do this exercise, I wrote

$\displaystyle ax^2 + bx + c \geq \frac{f(w+h) - f(w-h)}{(w+h)-(w-h)}\cdot(x-(w-h)) + f(w-h)$

$\hspace{2cm} = bx + 2awx - aw^2 + c + ah^2$

$x^2 - 2wx + w^2 - h^2 \leq 0\hspace{2cm}$as $a<0$

$\frac{2w-\sqrt{4w^2 -4(w^2 - h^2)}}2 \leq x \leq \frac{2w+\sqrt{4w^2 -4(w^2 - h^2)}}2$

$w-h\leq x\leq w+h$

However, what if $h<0$? Why does the solution fall apart?

Answer 1

It should be $\sqrt{4w^2-4(w^2-h^2)}=2|h|$ so that $w-|h|\le x\le w+|h|$.