Since we have a 2nd order ODE, our characteristic equation will be quadratic, and so we can use the quadratic formula to write the roots. But, if our characteristic equation is of the form:
[MATH]ar^2+br+c=0[/MATH]
Then, to express the solution in trigonometric form, we want to express the roots in complex form:
[MATH]r=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}[/MATH]
Do you see/understand why?