$\displaystyle \displaystyle e^{\lambda x}\neq 0$. Thus, the characteristic equation holds the only values that will make the equation 0.
$\displaystyle \lambda^2+a\lambda+b \ \mbox{is a quadratic equation which is solved from factoring or the quadratic equation} \ \lambda=\frac{-a\pm\sqrt{a^2-4(1)(b)}}{2(1)}$