#### Answer

a) For 2 real distinct negative roots, $b^2>4ac$ and $|\sqrt{b^2-4ac}|0$.
b) For 2 real distinct roots with opposite signs, $b^2>4ac$ and $b-\sqrt{b^2-4ac}=-(b+\sqrt{b^2-4ac})\Rightarrow{b}=0,\quad{a}<0,\quad{c}>0\quad{or}\quad a<0,\quad{c}>0$
c) For 2 real distinct positive roots, $b^2>4ac$ and $|\sqrt{b^2-4ac}|$

#### Work Step by Step

$$ay''+by'+cy=d$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$a{\lambda}^2+b{\lambda}+c=0$ ($\because$ equilibrium solution causes LHS=0)
$$\lambda_{1,2}=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$$
As with root analysis in quadratic equations, we will aim to deduce conditions in a, b and c by studying the discriminant $D=b^2-4ac$.
For 2 real distinct negative roots, $b^2>4ac$ and $|\sqrt{b^2-4ac}|0$.
For 2 real distinct roots with opposite signs, $b^2>4ac$ and $b-\sqrt{b^2-4ac}=-(b+\sqrt{b^2-4ac})\Rightarrow{b}=0,\quad{a}<0,\quad{c}>0\quad{or}\quad a<0,\quad{c}>0$
For 2 real distinct positive roots, $b^2>4ac$ and $|\sqrt{b^2-4ac}|$