Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector $v$ such that $$\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} {v} = k {v}.$$
I tried expanding but I could not collect anything from the equations I got. I have not learned eigenvalues and eigenvectors yet.
You haven't learned about eigenvalues yet, but this question is identical to asking for eigenvalues for the matrix $A = \begin{bmatrix} 2& 12 \\ 2 & -3 \end{bmatrix}$. There is a standard procedure for finding eigenvalues. For our particular matrix $A$, this goes as follows:
First we deform the problem into something that we can calculate. We see that for $k \in \mathbb{R}$, $\exists v \neq 0$ such that $Av = kv$ if and only if $\exists v \neq 0$ such that $(A - kI)v = 0$. But $\exists v \neq 0$ such that $(A - kI)v = 0$ if and only if $A - kI$ is not invertible (this is a fact about square matrices). $A - kI$ is invertible if and only if $det(A - kI) \neq 0$. And the great thing about this last point is that we can calculate whether $det(A - kI) \neq 0$ for any $k$, using the formula for determinant: \begin{align*} det(A - kI) = & \ det \left( \begin{bmatrix} 2 - k & 12 \\ 2 & -3 -k \end{bmatrix} \right) \\ = & (-3-k)(2-k) - 12*2 \\ = & k^2 + k - 30 \\ = & (k+6)(k-5) \end{align*} Therefore $det(A - k I ) = 0$ whenever $k = -6$ or $5$. We conclude that the values of $k$ such that there exists a nonzero $v \in \mathbb{R}^2$ such that $A v = kv$ are $-6$ and $5$.