Matrix Notation Form of Roots of a Quadratic Equation We know that the quadratic equation 
$$f(x)=ax^2+bx+c=0$$
 has roots 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$
Also, $f(x)$ can be written in matrix notation as follows:
$$f(x)=
\left(\begin{matrix}x&1\\\end{matrix}\right)
\left(\begin{matrix}a&\frac b2\\\frac b2&c\end{matrix}\right)
\left(\begin{matrix}x\\1\end{matrix}\right)=\mathbf{x^T Q x}$$
where the determinant of $\mathbf Q$ is $\left(ac-\frac {b^2}4\right)=-\frac 14\left(b^2-4ac\right)$, where coincidentally the familiar $(b^2-4ac)$ is the discriminant of the quadratic $f(x)$.  
Hence the roots of the quadratic $f(x)=0$ may be written as
$$x=-\frac b{2a}\pm \frac 1a\sqrt{-\det(\mathbf Q)}$$
This is equivalent to
$$\left(x+\frac b{2a}\right)^2=\frac {-\det(\mathbf Q)}{a^2}$$
Or in neater form, 
$$\left(ax+\frac b{2}\right)^2={-\det(\mathbf Q)}$$

Question
  Can the roots of $f(x)=0$ be derived and written completely in matrix notation, given the link between the determinant and discriminant as shown above?

 A: I have obtained some formula with a little changed notation comparing the notation used in the question.   
I've interchanged the  components of $\mathbf x$ (the new vector is denoted as  $\mathbf{ \hat{x}}$ ) and I've interchanged the entries on the main diagonal of $ \mathbf Q$ (new matrix is denoted as $ \mathbf M$ - interchanging entries on the diagonal for $2 \times 2$ matrices doesn't change the determinant) .   
Then $f(x)$ can be written down as
$$f(x)=
 \begin{bmatrix}1&x\\\end{bmatrix}  
 \begin{bmatrix}c&\frac b2\\\frac b2&a\end{bmatrix}  
 \begin{bmatrix}1\\x\end{bmatrix}  =\mathbf{\hat{x}^T M \hat{x}}=0$$
It can be calculated that
$ \begin{bmatrix}1 & x\\0 & 1\end{bmatrix} 
 \begin{bmatrix}c & \frac{b}{2}\\\frac{b}{2} & a\end{bmatrix} 
 \begin{bmatrix}1 & 0\\x & 1\end{bmatrix} =
 \begin{bmatrix}\frac{b x}{2} + c + x \left(a x + \frac{b}{2}\right) & a x + \frac{b}{2}\\a x + \frac{b}{2} & a\end{bmatrix}  =    \begin{bmatrix} 0 & a x + \frac{b}{2}\\a x + \frac{b}{2} & a \end{bmatrix} $
We know also that determinant of a matrix product is equal to the product of determinants of multiplied matrices.
Additionally we have $\det( \mathbf Q)=\det(\mathbf M)$. 
From this follows that 
$ \det( \mathbf Q)=-\left(   ax+ \frac{b}{2}\right)^2$.
A: Let us switch to homogeneous coordinates and use $2b$ instead of $b$ for convenience.
$$ax^2+2bxy+cy^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{p^TQp}=0.$$
Assuming $Q$ diagonalizable, we have (the transformation $P$ can be taken orthonormal)
$$\mathbf{p^TQp}=\mathbf{p^TP^TDPp}=\mathbf{q^TDq}=0.$$
The last expression is of the form
$$\lambda_0u^2+\lambda_1v^2=0$$ where the lambdas are the Eigenvalues of $\mathbf Q$. For the equation to have real solutions, the Eigenvalues must have opposite signs, or $\lambda_0\lambda_1\le0$, which is precisely $-\det(\mathbf Q)=b^2-ac\ge0$.
The conic factors as
$$\left(\sqrt{|\lambda_0|}u+\sqrt{|\lambda_1|}v\right)\left(\sqrt{|\lambda_0|}u-\sqrt{|\lambda_1|}v\right)=0$$
and the solutions are given by plugging $u=\mathbf{e_0p},v=\mathbf{e_1p}$ in the two factors, setting $y=1$ and solving the linear equations for $x$.
