Given a real-valued matrix $$ M = \begin{pmatrix} a & b \\ b & c \end{pmatrix} $$ with $a<0 <b$ and $a c < b^2$, how can I characterize the values $x = (x_1,x_2)$ that satisfy $x^\top M x > 0$?
The eigenvalues of $M$ are $$ \lambda_\pm = \frac{a+c}2 \pm \sqrt{\Big(\frac{a-c}{2}\Big)^2+b^2}. $$ The condition $ac<b^2$ implies $\lambda_- < 0 < \lambda_+,$ hence $M$ is not positive definite or negative definite.
$M$ is simetric and so there exist $U$ such that $M=U^{t}\begin{pmatrix} \lambda_{+} & 0 \\ 0 & \lambda_{-} \end{pmatrix}U,\ UU^{t}=U^{t}U=I$. Since $x^{t}Mx=(Ux)^{t}\begin{pmatrix} \lambda_{+} & 0 \\ 0 & \lambda_{-} \end{pmatrix}(Ux)$ and $\lambda_- < 0 < \lambda_+,$, the $(x_{1},x_{2})$ shearched are precisly those such that $Ux=(\mu_1,\mu_2)^{t}$ satisfies $\lambda_{}\mu_{1}^{2}+\lambda_{-}\mu_{2}^{2}>0$. I left the details to you.