= Solution
A <reduced positive definite binary quadratic form> $f=[a,b,c]=ax^2+bxy+cy^2$ satisfies
$$
|b|\leq a\leq c,
$$
with $b\geq0$ when $|b|=a$ or $a=c$. Its <discriminant of a binary quadratic form> is $d=b^2-4ac<0$. The first displayed inequality already gives $|b|\leq a$, while $c\geq a$ gives
$$
|d|=4ac-b^2\geq4a^2-a^2=3a^2.
$$
Taking the positive <square root> yields
$$
\boxed{|b|\leq a\leq\sqrt{|d|/3}.}
$$
Moreover, $d=b^2-4ac\equiv b^2\equiv b\pmod2$, so $b\equiv d\pmod2$. For fixed negative $d$, the displayed bounds leave only <finite set>[finitely many] possible <integers> $a$ and $b$, and then $c=(b^2-d)/(4a)$ is uniquely determined. Hence only finitely many reduced forms have discriminant $d$.
For $d=-15$, the bound gives $a\leq\sqrt5$, so $a\in\{1,2\}$. The <modular congruence> condition makes $b$ <odd number>[odd]. If $a=1$, then $b=\pm1$ and $c=4$; the boundary convention $|b|=a\Rightarrow b\geq0$ retains $[1,1,4]$. If $a=2$, then $b=\pm1$ and $c=2$; the boundary convention $a=c\Rightarrow b\geq0$ retains $[2,1,2]$. Thus the complete list is
$$
\boxed{[1,1,4],\qquad [2,1,2].}
$$
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