<s>[INST] Small clarification Spivak Chapter 1, 18(b) 
Suppose that ${b}^{2} - 4c < 0$. Show that there are no numbers $x$ satisfying ${x}^{2} + bx + c = 0$; in fact, ${x}^{2} + bx + c > 0$ for all $x$. Hint; Complete the square.

There is one answer here.
Solution (per Spivak)

We have
${x}^{2} + bx + c = \left (x + \frac{b}{2} \right)^{2} + \left(c - \frac{b^{2}}{4} \right) \ge c - \frac{b^{2}}{4} $

Analysis
$\left (x + \frac{b}{2} \right)^{2} $ is  $\ge 0$ as any real number $a$, $a^{2} \ge 0$
$\left(c - \frac{b^{2}}{4} \right)$ See question below.
Spivak now states

but $c - \frac{b^{2}}{4} > 0$, so ${x}^{2} + bx + c > 0$ for all $x$

Question
Ignoring the conclusion (included for completeness), what allows him to assert $c - \frac{b^{2}}{4} > 0$?
 [/INST] Since the assumption (given condition) was,
$$b^2-4c<0\implies c-\frac{b^2}{4}>0.$$
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