I have tried to derive the quadratic formula using methods from linear algebra, and I wanted to check if my derivation is indeed valid and correct.
I started of with a pair of simultaneous equations with $2$ unknowns: $$\begin{align*} x^2-y &= 0\\ bx + ay+c&=0 \end{align*} $$
What I have done is I have replaced the $x^2$ term with $y$, hence the first equation. Normally I would have put $ay+bx+c=0$ as the first equation, but for the sake of making all the terms align with each other nicely, I have put it as $bx+ay+c=0$.
To solve these simultaneous equations, I have used matrices. When we use matrices, we get the following: $${\begin{bmatrix}x & -1\\b & a\end{bmatrix}}^{-1}\begin{bmatrix}0\\-c\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-\frac{c}{ax+b}\\-\frac{cx}{ax+b}\end{bmatrix}$$
As mentioned previously, $y=x^2$, which means that $(-\frac{c}{ax+b})^2=-\frac{cx}{ax+b}$. We will solve this equation.
I'll leave out the steps for solving it, since it is just a matter of rearranging to make $x$ the subject and both WolframAlpha and Symbolab believe it to be correct. However, what is interesting is that we get $x = \frac{-b+\sqrt{b^2-4ac}}{2a}$, $x = \frac{-b-\sqrt{b^2-4ac}}{2a}$ , which can be expressed succinctly as $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. We have successfully derived the quadratic formula, but what I want to know is if my method is correct and if there is any way in which I could improve my approach.