Rationally Odd

Let us assume the quadratic has rational roots. Thus it's discriminant is a perfect square of a natural number.

$\therefore b^2-4ac=k^2 (Let)$

Since $b$ is odd, $b^2$ is odd.

$\therefore k^2$ is odd $\Rightarrow k$ is odd

Rearranging the above relation gives us,

$b^2-k^2=4ac$

Since $b$ and $k$ are odd integers,

$8\mid b^2-k^2$

Also $4 \mid 4ac$ . But since $a$ and $c$ are odd integers, $8 \nmid 4ac$

Which is a contradiction to our assumption. Thus, no quadratic with only odd integral coefficients has rational roots.