Well. . .
There's several problems here.
First, notice:
$$(x+d)(x-e)=x^2-ex+dx-ed=x^2+(d-e)x-ed$$
Where in the world are you going to get a leading coefficient? In other words, this method doesn't suffice without some manipulations of the standard form.
That manipulation is:
$$\frac{1}{a}\left(ax^2+bx+c\right)=x^2+px+q \quad \quad p=\frac{b}{a},q=\frac{c}{a}$$
The second issue is that it can't be generally figured out what $d$ and $e$ are. $d$, by definition, is the opposite of the first root of the equation. $e$, by definition, is the second root. However, the relationships between $p$ and $q$ and $d$ and $e$ are:
$$d-e=p \quad -ed=q$$
This information doesn't allow us to determine $d$ and $e$ in terms of the coefficients $p$ and $q$. There's also the common sensical fact, that like I said, you have to know the roots to know $e$ and $d$.
But the biggest issue is what you want. You cannot get rid of the imaginary solutions. They will always be solutions to particular quadratics. It is impossible to "get rid of them." For example, notice:
$$x^2+1=0$$
There are no solutions to this equation in $\mathbb{R}$. The only solutions are in $\mathbb{C}$. They are $i$ and $-i$. Now, let me show why they have to be those particular quantities in a very simply way:
$$\begin{align}
x^2+1&=(x-r_1)(x-r_2)\\
&=(x-i)(x+i)\\
&=x(x+i)-i(x+i)\\
&=x^2+ix-ix-i^2\\
&=x^2-(-1)\\
&=x^2+1
\end{align}$$
Do you see how those solutions have to be $i$ and $-i$? This applies to all quadratics. Their roots are whatever makes the following relation true and the fundamental relation (that is, that the quadratic is equal to $0$) true. I hope this makes sense.
Addendum: It may help to see precisely where "that formula" comes from. It is derived directly from the standard form equation.
First, make this observation:
$$(x-v)^2=x^2-2vx+v^2$$
You can turn the manipulated form into a square:
$$\begin{align}
x^2+px+q&=0\\
x^2+px&=-q\\
x^2+px+\left(\frac{p}{2}\right)^2&=-q+\left(\frac{p}{2}\right)^2\\
\left(x+\frac{p}{2}\right)^2&=-q+\left(\frac{p}{2}\right)^2=\frac{-4q+p^2}{4}\\
x+\frac{p}{2}&=\pm \sqrt{\frac{-4q+p^2}{4}}=\frac{1}{2}\sqrt{-4q+p^2}\\
x&=\pm \frac{1}{2}\sqrt{-4q+p^2}-\frac{p}{2}=\frac{-p\pm\sqrt{-4q+p^2}}{2}
\end{align}$$
Substituting back in for $p$ and $q$ gives us:
$$\begin{align}
x&=\frac{-\frac{b}{a}\pm \sqrt{-4\frac{c}{a}+\left(\frac{b}{a}\right)^2}}{2}\\
&=\frac{-b \pm a\sqrt{-4\frac{c}{a}+\frac{b^2}{a^2}}}{2a}\\
&=\frac{-b \pm a\sqrt{\frac{b^2-4ac}{a^2}}}{2a}\\
&=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\end{align}$$
Also, André Nicolas provides an even more simple derivation of the quadratic formula in this question: Why can ALL quadratic equations be solved by the quadratic formula?
Let $u=A^{-1}a$, then:
\begin{align}⟨a−Ax,B(a−Ax)⟩&=(-1)^2⟨A(x-A^{-1}a),BA(x-A^{-1}a)⟩ \\
&=(x-u)^TA^TBA(x-u) \\
&=(x-u)^T\frac 12(A^TBA+(A^TBA)^T)(x-u) \\
&=(x-u)^T\frac 12(A^T(B+B^T)A)(x-u) \\
&=(x-u)^TUDU^T(x-u) \\
&\le t
\end{align}
This is a quadratic form like an ellipsoid, translated by $u$, transformed by the orthogonal matrix $U$, and identified by the diagonal matrix $D$.
Best Answer
Using this approach you can derive a formula for solving quadratic equations. Let $a x^2 + b x - c$ be a polynomial with ($a \neq 0$), then:
$$a x^2 + b x = -c \iff a(x^2 + \frac{b}{a}x) = c$$ $$\iff a(x+\frac{b}{a}x +\frac{b}{2a}) = c + \frac{b}{2a}$$ $$\iff a(x+\frac{b}{2a})^2 = c + \frac{b}{2a}$$ $$\iff \pm\sqrt{a}(x+\frac{b}{2a}) = \sqrt{c+\frac{b}{2a}}$$ $$\iff x = \frac{\pm\sqrt{c+\frac{b}{2a}}}{\sqrt{a}} - \frac{b}{2a} $$