Solving a quadratic equation with parameters $$\frac { x+2 }{ a }- \frac { 2-a }{ x } =2$$
Steps I took:
$$\frac { x(x+2) }{ ax } -\frac { a(2-a) }{ ax } =\frac { 2ax }{ ax } $$
$$x^2+2x-2a+a^2=2ax$$
$$x^{ 2 }+2x-2a+a^{ 2 }-2ax=0$$
I get stuck at this point because I have no clue how to factor an equation in this form. Hints a more appreciated than the actual answer
 A: Notice, we have $$\frac{x+2}{a}-\frac{2-a}{x}=2$$
$$\frac{x(x+2)-a(2-a)}{ax}=2ax$$  $$x^2+2x-a(2-a)=2ax$$
$$x^2-2(a-1)x+a(a-2)=0$$ $$x=\frac{2(a-1)\pm\sqrt{4(a-1)^2-4a(1)(a-2)}}{2(1)}$$
$$=\frac{2(a-1)\pm\sqrt{4(a-1)^2-4a(1)(a-2)}}{2(1)}=\frac{2(a-1)\pm 2}{2(1)}$$ 
$$=\frac{2(a-1)\pm 2}{2}$$ 
 $$\iff x=\frac{2(a-1)+2}{2}=a$$
$$\iff x=\frac{2(a-1)-2}{2}=a-2$$
Hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=a,\ (a-2)}}$$
A: $$x^2+(2-2a)x+a^2-2a=0$$
use the quadratic formula according to the following form
$$x^2+Ax+B=0$$
$$x=-0.5A\pm\sqrt{0.25A^2-B}$$
hence
$$x=(a-1)\pm\sqrt{(a-1)^2-a^2+2a}$$
$$x=(a-1)\pm\sqrt{a^2-2a+1-a^2+2a}$$
$$x=(a-1)\pm{1}$$
A: Hint: this is not a general method but you might spot $$x^2+2x-2a+a^2-2ax=(x^2-2ax+a^2)+2x-2a=(x-a)^2+2(x-a)$$

Also be careful about cases where $ax=0$ since then you can't say, for example, that $\cfrac {2ax}{ax}=2$
A: You have $x^2+2x−2a+a^2−2ax=0$, so first write this in standard form $$x^2+(2-2a)x+(a^2−2a)=0.$$  
Remember the standard binomial identities $(x\pm a)^2=x^2\pm2ax+a^2$, and from the $x$ term in your equation notice
$$(x+(1-a))^2=x^2+(2−2a)x+(1-a)^2,$$
and use this to rewrite your equation as
$$\left[(x+(1-a))^2-(1-a)^2\right]+a^2-2a=0.$$
Now simplify
$$(x+(1-a))^2=(1-a)^2-a^2+2a$$
$$(x+(1-a))^2=1-2a+a^2-a^2+2a$$
$$(x+(1-a))^2=1$$
and solve by taking both square roots
$$x+1-a=\pm1$$
Giving
$x=a$, or $a-2$.
This is called completing the square and is where the quadratic formula comes from.  I think it is both clearer and safer than than using the formula.  I have put in lots of steps to make it clear, but most of them you would do in your head when comfortable with the approach.
A: just use $$x^{ 2 }+(2-2a)x+(-2a+a^{ 2 })=0$$
A: HINT : 
$$x^2+2x-2a+a^2-2ax=0$$
$$\Rightarrow x^2+(2-2a)x+a^2-2a=0$$
$$\Rightarrow x^2+\color{green}{(2-2a)}x+\color{red}{a}\color{blue}{(a-2)}=0$$
Here, note that $-\color{red}{a}-\color{blue}{(a-2)}=\color{green}{2-2a}$.
