Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible

Let
\[A=\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}.\]
Then $\frac{1}{a}R_1+R_3 \rightarrow R_3$ gives 
\[\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & x & x^2+\frac{c}{a} \end{pmatrix},\]
and $-\frac{x}{a}R_2+R_3 \rightarrow R_3$ gives 
\[\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & 0 & \frac{b}{a}x+x^2+\frac{c}{a} \end{pmatrix}.\]
Hence
\[\text{det}(A)=a(\frac{b}{a}x+x^2+\frac{c}{a})=0\]
\[\Rightarrow ax^2+b x+c=0\]
\[\Rightarrow x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]
Hence the matrix $A$ is invertible if and only if
\[x\neq\frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]