Consider the matrix
$$ G = \begin{pmatrix} 3 & 1 & -2 \\ 1 & 2 & 0 \\ -2 & 0 & -3 \\ \end{pmatrix}$$
and the quadratic form $q: \mathbb{R}^3 \to \mathbb{R}$, given by $q(v) = v^t G v$.
I now want to find $r, r_+ \in \mathbb{N}_0$ with $r + r_+ ≤ 3$, so that q is equivalent to
$$Q: \mathbb{R}^3 \to \mathbb{R}, (x_1, x_2, x_3) \mapsto x_1^2 + ... + x_{r_+}^2 - x_{r_+ + 1}^2 - ... - x_r^2$$
using the definition that two quadratic forms $a: V \to K$, $b: W \to K$ are equivalent if there exists a linear isomorphism $\alpha: V \to W$ so that $a = b \circ \alpha$.
What I've done so far: I picked a random element $v = \begin{pmatrix} x_1 \\ x_2 \\x_3 \end{pmatrix} \in \mathbb{R}^3$ and worked out that $q(v) = 3x_1^2 + 2 x_2^2 + 2x_1 x_2 - 3 x_3^2 - 4 x_1 x_3$. Now I think we must choose $\alpha$ so, that we somehow get rid of the terms $2x_1 x_2$ and $4x_1 x_3$, because the square factors seem fine to me. I don't know how to work out $\alpha$. Thanks in advance.
There's a systematic way to do this.
First, take any $v_1$ such that $q(v_1)\neq 0$. For example, $v_1 = \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}$ works.
Next, we find $v_2$ such that $v_2^TGv_1=v_2^T\begin{pmatrix} 3 \\ 1 \\-2 \end{pmatrix}= 0$ and $q(v_2)\neq 0$. For example, $v_2=\begin{pmatrix} 1 \\ 1 \\2 \end{pmatrix}$ works.
Next, find $v_3$ such that $v_3^TGv_1=v_3^TGv_2=0$ and $q(v_3)\neq 0$. For example, $v_3=\begin{pmatrix} -2 \\ 24 \\9 \end{pmatrix}$ works.
Finally scale each $v_i$ so that $|q(v_i)|=1$. It turns out that $q(v_1)=3$, $q(v_2)=-13$, $q(v_3)=897$. After scaling and reordering, we get
$u_1=\frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}, u_2=\frac{1}{\sqrt{897}}\begin{pmatrix} -2 \\ 24 \\9 \end{pmatrix},u_3=\frac{1}{\sqrt{13}}\begin{pmatrix} 1 \\ 1 \\2 \end{pmatrix}$.
By construction, the $u_i$ are a basis for $\Bbb{R}^3$ with the property that $q(\lambda_1u_1+\lambda_2u_2+\lambda_3u_3)=\lambda_1^2+\lambda_2^2-\lambda_3^2$. Using your notation, we have $r_+=2$ and $r=1$.
Now recall that if $A,B$ are matrices representing a quadratic form $q$ with respect to bases $\mathcal{B}_1$, $\mathcal{B}_2$ respectively, and if $P$ is the change of basis matrix from $\mathcal{B}_1$ to $\mathcal{B}_2$, then $B=P^TAP$. In our case, if $P=\begin{pmatrix} u_1 & u_2 &u_3 \end{pmatrix}$, then $P^TGP=\begin{pmatrix} 1 &0& 0 \\ 0 & 1 & 0 \\0 & 0 & -1 \end{pmatrix}$. So, the desired isomorphism $\alpha$ is that defined by $P$, i.e. $\alpha:\Bbb{R}^3\rightarrow\Bbb{R}^3$, $\alpha(e_i)=u_i$ for $i=1,2,3$, and extending linearly. (Here the $e_i$ are just the standard basis vectors for $\Bbb{R}^3$.)
Remark The $u_i$ were arbitrarily chosen; you can just as well make other choices. However, $r_+$ and $r$ are unique and independent of this choice.