X^2 - 5x + 6 = 0

No. This is how maths work. You just do stuff and hope it all adds up to what you want, no matter how arbitrary it may seem

You don't need to multiply by 4a. See pic related, the superior proof of the quadratic formula.

here's a more intuitive way, using the method of completing the square:
[math] \begin{align*}
ax^2 + bx + c &= 0 \\
x^2 + \frac{b}{a}x &= -\frac{c}{a} \\
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 &= -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \\
\left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{c}{a} \\
\left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \\
x + \frac{b}{2a} &= \frac{\pm\sqrt{b^2 - 4ac}}{2a} \\
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*} [/math]

((x + y)^2 - 4xy)/(x - y)
I'm assuming you meant to write it like so^
x = y

(((x + y)^2 - 4xy)/x) - y

x = +or- y(sqrt(5) + 3)/2

idk how this nigga got x = 2 or x = 3; he probably shouldn't be doing this shit in his head

Factorise: x^2 - 5x + 6 = 0
So: (x-2)(x-3)=0
X=2, X=3 makes zero
Thus X=2 and X=3 are the solutions

>x^2 - 5x + 6

I'm talking about OP's second question:>((x + y)^2 - 4xy)/x - y

>What about this?

>idk how this nigga got x = 2 or x = 3
Really?
Read the first fucking post, faggot

MUH DICK
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I'm not talking about the question original post brainlet

was in response to this:not the original post

X^2-5x+6=0
X(x-5)= -6
X=-6
X-5=-6
X=-1