Structure of the general solution from the characteristic roots

Try the ansatz $y=e^{rx}$. Then $y'=re^{rx}$ and $y''=r^2e^{rx}$; substituting into $ay'' + by' + cy = 0$ gives $(ar^2+br+c)e^{rx}=0$. Since $e^{rx}$ is never zero, this forces the characteristic equation $ar^2+br+c=0$. By the quadratic formula, $r = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Case $\Delta>0$: the two roots $r_1\neq r_2$ give solutions $e^{r_1x}$ and $e^{r_2x}$. Their Wronskian is $(r_2-r_1)e^{(r_1+r_2)x}\neq 0$, so by the superposition theorem the general solution is $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$.

Case $\Delta=0$: there is one repeated root $r=-b/(2a)$, giving only one exponential solution $e^{rx}$. To find a second independent solution, try $y_2=v(x)e^{rx}$ (reduction of order) and substitute into the equation; because $r$ is a double root of $ar^2+br+c$, the terms in $v$ and $v'$ cancel, leaving $av''e^{rx}=0$, i.e. $v''=0$, so $v=c_1+c_2x$. This yields the second solution $xe^{rx}$, and the general solution $y = (c_1 + c_2 x) e^{r x}$.

Case $\Delta<0$: the roots are complex conjugates $r = \alpha \pm i\beta$ with $\alpha=-b/(2a)$, $\beta=\sqrt{-\Delta}/(2a)$. Euler's formula $e^{i\beta x}=\cos(\beta x)+i\sin(\beta x)$ turns the complex solutions $e^{(\alpha\pm i\beta)x}$ into two independent real solutions $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ (by taking the real and imaginary parts, which are themselves solutions since $L$ has real coefficients), giving the general solution $y = e^{\alpha x}\left(c_1 \cos(\beta x) + c_2 \sin(\beta x)\right)$.