$abc$-formula: $D >= 0$ or $D > 0$?

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I don't really understand when to put $>$ or $<$ , $>=$ or $<=$ when working with the $abc$-formula.

This is the $abc$-formula:

We have a quadratic equation: $$ax^2+bx+c$$ Now for the $abc$-formula: $$D = (b)^2 -4\times a c$$ $$D > 0 \text{ two solutions, } D < 0 \text{ no solutions} D = 0 \text{ one solution}$$ $$x = \frac{-b\pm \sqrt{D}}{2\times a}$$


Now, for example, the following problem is given:

$$f_{p}(x) = x^3 +5x^2 +px$$

For what value of $p$ has $f_{p}$ no extreme value? So I would presume we would need $D < 0$

Here's how I worked out this problem:

$$f'_{p}(x) = 3x^2 +10x+p = 0$$ $$D = (10)^2 -4 \times 3 \times p$$ $$100- 12 p $$

$D < 0$ gives us no solution so:

$$100-12p < 0$$ $$-12p < -100$$ $$p > \frac{-100}{-12}$$ $$p > \frac{25}{3} $$

I solved it correctly but my book says it has to be the following:

$$100-12p \leqslant 0$$ $$-12p \leqslant -100$$ $$p \geqslant \frac{-100}{-12}$$ $$p \geqslant \frac{25}{3} $$

(If you need more information please ask and I'll supply.)

All help is highly appreciated.

Cro-Magnon