\[
y''-2y'-2y=0
\]
\[
y''-2y'-2y=0
\]
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation
\[
r^2-2r-2=0
\]
We use the quadratic formula which is
\[
r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
Hence,
\[
\left\{
\begin{array}{c}
r_1=1+\sqrt3\\
r_2=1-\sqrt3\\
\end{array}
\right.
\]
Since the general solution has the form of
\[
y=c_1e^{r_1t}+c_2e^{r_2t}
\]
Therefore, the general solution of the given differential equation is
\[
y=c_1e^{(1+\sqrt3)t}+c_2e^{(1-\sqrt3)t}
\]
