Find All Values of x such that the Matrix is Invertible

Given any constants a,b,c$a,b,c$ where a≠0$a\ne 0$, find all values of x$x$ such that the matrix A$A$ is invertible if

A=⎡⎣⎢10−1/a0axc−bx2⎤⎦⎥.

det(A)=1∣∣∣ax−bx2∣∣∣+c∣∣∣0−1/aax∣∣∣=1(ax2+bx)+c(0+1)=ax2+bx+c.$$\begin{array}{}\end{array}$$det(A)=1|a−bxx2|+c|0a−1/ax|=1(ax2+bx)+c(0+1)=ax2+bx+c.

In this manner det(A)≠0 when ax2+bx+c≠0. We know by the quadratic equation that ax2+bx+c=0 definitely when 

x=−b±b2−4ac−−−−−−−√2.