Answer
$$
y^{\prime \prime}+4 y^{\prime}+6.25 y=0
$$
The general solution of that equation is given by
$$ y(t) =e^{-2t}( c_{1} \cos \frac{3t}{2} +c_{2} \sin \frac{3t}{2}) $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}+4 y^{\prime}+6.25 y=0 \quad\quad\quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^2+4r+6.25=100r^2+400r+625=0
$$
so its roots are
$$
r_{1,\:2}=\frac{-400\pm \sqrt{400^2-4\cdot \:100\cdot \:625}}{2\cdot \:100}
$$
Thus the possible values of $r$ are
$$r_{1}=-2+i\frac{3}{2} ,\quad r_{2}=-2-i\frac{3}{2}.$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-2+i\frac{3}{2})t} =e^{-2t}(\cos \frac{3t}{2} + i \sin \frac{3t}{2})
$$
and
$$
y_{2}(t) = e^{(-2-i\frac{3}{2})t} =e^{-2t}(\cos \frac{3t}{2} - i \sin \frac{3t}{2})
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-2t}( c_{1} \cos \frac{3t}{2} +c_{2} \sin \frac{3t}{2})
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
