Answer
$$a= 2,\ \ a=-6$$
Work Step by Step
Given $$y^{\prime \prime}+4 y^{\prime}-12 y=0$$
Since $y=e^{ax}$ is a solution and
\begin{align*}
y'&=ae^{ax}\\
y''&=a^2e^{ax}
\end{align*}
Hence
\begin{align*}
y^{\prime \prime}+4 y^{\prime}-12 y&=0\\
a^2e^{ax}+4ae^{ax}-12e^{ax}&=0\\
[a^2 +4a -12]e^{ax}&=0\\
(a-2)(a+6)e^{ax}&=0
\end{align*}
Then $$a= 2,\ \ a=-6$$
