Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero)

Essentially the same algebraic manipulations used in the second part of your question will give a proof without aiming for a contradiction.

Indeed, these manipulations will give you that $x^2 + bx + c = (x + \frac{b}{2})^2 - \frac{b^2 - 4c}{4}$. This is strictly positive since $(x + \frac{b}{2})^2 \geq 0$ and $\frac{b^2 - 4c}{4} < 0$. In particular, $x^2 + bx + c > 0$ for every $x$ which completes the proof.

Given that $b^2-4c<0$,

$$x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge-\frac{b^2}{4}+c=-\frac{1}{4}\left(b^2-4c\right)>0$$

From here: $(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$ we get $b^2-4ac \geq 0$ and not $>$. So if $b^2-4c<0$ there is no real solution. So else from that $>$ your conclusion is correct.