Consider the arbitrary quadratic equation $ax^2 + bx + c = 0$ for real numbers $a$, $b$, $c$ with $a\neq0$. By the fundamental theorem of algebra this equation has two roots $x_1$, $x_2$, over the complex numbers. Using the factor theorem gives the factorization $ax^2 + bx + c = a(x-x_1)(x-x_2) .$ Expanding out the right-hand side gives $ax^2 - a(x_1+x_2)x + ax_1x_2 ,$ and equating coefficients with the original expression gives the following system of equations: \begin{align*} b &= - a(x_1+x_2) \\ c &= ax_1x_2 \end{align*} And now it is non-obvious how to proceed and solve for $x_1$, $x_2$.
It turns out that it is fruitful to introduce the new variable $y_1$ by the definition $y_1 = a(x_1-x_2) .$ Then the following identity is satisfied: $y_1^2 = b^2 - 4ac .$ This can be found using Maple, for example.
Now $y_1$ can be found by taking a square root. In this case, either square root will work, although the choice will affect which root is named $x_1$ and which is named $x_2$.
Once $y_1$ has been determined, one can solve the linear system \begin{align*} b &= - a(x_1+x_2) \\ y_1 &= a(x_1-x_2) \end{align*} for $x_1$ and $x_2$. The solution is given by \begin{align*} x_1 &= (-b+y_1)/(2a) \\ x_2 &= (-b-y_1)/(2a) \end{align*} or more compactly by $x = \frac{-b\pm y_1}{2a} ,$ which is of course the familiar formula.