Solving simultaneous equations involving a quadratic I have the question 
Solve the simultaneous equation  pair
$$x^2 + y^2 = 25\tag1$$
$$2x - y = 5\tag2$$

I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.
I get $x^2 + (2x -5)^2 = 25$
When I expand the brackets I get the equation 
$$x^2+4x^2-20 =0\tag3$$
However, when I checked the solutions the equation should simplify to $x^2 - 4x = 0$ and I do not understand how this is achieved.

$$x^2 + y^2 = 25 $$
$$2x - y = 5 \Leftrightarrow y=2x-5$$
$$ x^2 + (2x-5)^2 = 25 $$
$$ x^2 +4x^2 -20x +25 = 25 $$
$$ 5x^2 -20x=0 $$
