Quadratic equation Suppose that $a+b = 2m_{1}$ and $ab = 4m_{1}^{2}-3m_{2}$. Why is the quadratic equation $$y^{2}-2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$ instead of $$y^{2}+2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$
In other words, why is it $-2m_{1}y$ instead of $2m_{1}y$ in the second term?
 A: Because, if a quadratic equation has two roots $x_1$ and $x_2$, then one has
$$
x^2+bx+c=(x-x_1)(x-x_2).
$$
Comparing the coefficients at the equal powers yields
$$
x_1+x_2=-b,\quad x_1 x_2=c,
$$
note the minus sign.
A: if you have equation like this
 $x^2+b*x+c$
then it is called  standard quadratic equation and by Vieta's Theorem,if $x_1$ and $x_2$ are roots of this equation then
$x_1+x_2=-b$
and  
$x_1*x_2=c$
EDITED:
so  see it by example
we know  that if $x_1$,$x_2$  are roots then
$(x^2+b*x+c)$=(x-$x_1$)*(x-$x_2$)
then let us   make operations on the right side,we have 
 $x^2-x*(x_1+x_2)+x_1*x_2$
now if we look at  $x^2+b*x+c$  
we can see  that $x_1+x_2=-b$ (to make  -  sign  positive) and  $x_1*x_2=c$
A: $a+b = x$   and $ab = y$ 
$$(a+b)^2 = x^2$$
$$a^2+2ab+b^2 = x^2$$ 
$$a^2+2ab+(x-a)^2 = x^2$$ 
$$a^2+2y+x^2-2xa+a^2 =x^2$$
$$2a^2+2y-2xa =x^2-x^2=0$$ 
$$2(a^2-xa+y) =0$$ 
Equation $(1)$
$$a^2-xa+y =0$$ 
if you put $a=x-b$  then
$$(x-b)^2-x(x-b)+y =0$$ 
$$x^2-2xb+b^2-x^2+xb+y =0$$
Equation $(2)$ 
$$b^2-xb+y =0$$
As you see Equation $(1)$ and Equation $(2)$ are the same Quadratic equation . $a$ and $b$ are roots of the Quadratic equation. As you see $x$ must be minus  $y$ must be plus while writing the quadratic equation.
