An alternative way to solve this problem is by using the [Wronskian][1]. Put
\begin{align*}
f(t) &= 1 + 2\,t+ t^2 & g(t) &= 3-9\,t^2 & h(t) &= 1 + 4\,t + 5\,t^2
\end{align*}
and define
$$
W(t)=
\begin{bmatrix}
f(t) & g(t) & h(t) \\
f^\prime(t) &g^\prime(t) & h^\prime(t) \\
f^{\prime\prime}(t) & g^{\prime\prime}(t) & h^{\prime\prime}(t)
\end{bmatrix}
=
\begin{bmatrix}
1+2\,t+t^2 & 3-9\,t^2 & 1+4\,t+5\,t^2 \\
2+2\,t & -18\,t & 4+10\,t \\
2 & -18 & 10
\end{bmatrix}
$$
If there exists a $t_0$ such that $\det W(t_0)\neq 0$, then $\{f,g,h\}$ is [linearly independent][2]. Since each of $f$, $g$, and $h$ are [analytic][3], if $\det W(t)=0$ for all $t\in\Bbb R$, then $\{f,g,h\}$ is linearly dependent. 

Now, note that adding $\DeclareMathOperator{Col}{Col}9\cdot\Col_1$ to $\Col_2$ and subtracting $5\cdot\Col_1$ from $\Col_3$ gives
$$
\det W(t)
=
\begin{vmatrix}
1+2\,t+t^2 &12+18\,t& -4-6\,t \\
2+2\,t&18&-6 \\
2&0&0
\end{vmatrix}
$$
Then adding $3\cdot\Col_3$ to $\Col_2$ gives
$$
\det W(t)
=
\begin{vmatrix}
1+2\,t+t^2 & 0 & -4-6\,t \\
2+2\,t & 0 & -6 \\
2 & 0 & 0
\end{vmatrix}
=
0
$$
This implies that $\{f,g,h\}$ is linearly dependent.


 [1]: https://en.wikipedia.org/wiki/Wronskian
 [2]: https://en.wikipedia.org/wiki/Wronskian#The_Wronskian_and_linear_independence
 [3]: https://en.wikipedia.org/wiki/Analytic_function