Vieta's formulas for quadratic equation problem I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it.
Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divide roots by $a$.
For example:
$$-6x^2+7x+5=0$$
I solve
$$
y^2 + 7y -30 = 0\\
y_1=-10\\
y_2=3
$$
And then divide the roots:
$$
x_1=\frac{-10}{-6}=\frac{5}{3}\\
x_2=\frac{3}{-6}=-\frac{1}{2}
$$
Which gives me a correct answer. But I want to know why it's so. For now what I figured out is only that:
$$
\text{for } a\neq 0 \text{ :}\\
ax^2 + bx + c = 0 \Leftrightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0\\
x_1 + x_2 = -\frac{b}{a} \Leftrightarrow a(x_1 + x_2) = -b\\
x_1 \cdot x_2 = \frac{c}{a} \Leftrightarrow a(x_1 \cdot x_2) = c\\
$$
$$
y^2 + by + ca = 0\\
y_1 + y_1 = -b\\
y_1 \cdot y_2 = ca \Leftrightarrow c = \frac{y_1 \cdot y_2}{a}\\
$$
$$
a(x_1 + x_2) = y_1 + y_1 \Leftrightarrow x_1 + x_2 = \frac{y_1}{a} + \frac{y_2}{a}\\
a(x_1 \cdot x_2) = \frac{y_1 \cdot y_2}{a} \Leftrightarrow x_1 \cdot x_2 = \frac{y_1}{a} \cdot \frac{y_2}{a}\\
$$
Any ideas?

Take the original equation $ax^2+bx+c=0$ then multiply by $a$ to obtain $$a^2x^2+abx+ac=0$$Now set $y=ax$ so that $$y^2+by+ac=0$$ and you are done.
