To determine this, you need to use the steps from above. The $x$, $y$ and $z$ terms needs to be rewritten as complete squares.

Items 1 and 2.

These steps can be done at the same time, as the goal is to complete the squares. To make $x^2+4x$ a square, you need to divide 4 (the number in front of $x$) by 2, and add the square of this to both sides of the equation. You get:

$$\begin{array}{llll}\hfill x^2+4x+{\left(\frac{4}{2}\right)}^2+\cdots & =\cdots +{\left(\frac{4}{2}\right)}^2& \hfill & \\ \hfill {\left(x+\frac{4}{2}\right)}^2+\cdots & =\cdots +{\left(\frac{4}{2}\right)}^2& \hfill & \end{array}$$

To make $y^2-2y$ a square, add ${\left(\frac{-2}{2}\right)}^2$ on both sides. You get:

$$\begin{array}{llll}\hfill y^2-2y+{\left(\frac{-2}{2}\right)}^2+\cdots & =\cdots +{\left(\frac{-2}{2}\right)}^2& \hfill & \\ \hfill {\left(y+\frac{-2}{2}\right)}^2+\cdots & =\cdots +{\left(\frac{-2}{2}\right)}^2& \hfill & \end{array}$$

You see that $z^2$ already is a complete square, so you can leave it alone.

Item 3.

Now, all you have to do is combine the calculations above:

$$\begin{array}{llll}\hfill x^2+4x+y^2-2y+z^2& =4& \hfill & \\ \hfill {\left(x+\frac{4}{2}\right)}^2+{\left(y+\frac{-2}{2}\right)}^2+z^2& =4+{\left(\frac{4}{2}\right)}^2& \hfill & \\ \hfill & +{\left(\frac{-2}{2}\right)}^2& \hfill & \\ \hfill {\left(x+2\right)}^2+{\left(y-1\right)}^2+z^2& =4+4+1& \hfill & \\ \hfill {\left(x+2\right)}^2+{\left(y-1\right)}^2+z^2& =3^2& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill x^2+4x+y^2-2y+z^2& =4& \hfill & \\ \hfill {\left(x+\frac{4}{2}\right)}^2+{\left(y+\frac{-2}{2}\right)}^2+z^2& =4+{\left(\frac{4}{2}\right)}^2+{\left(\frac{-2}{2}\right)}^2& \hfill & \\ \hfill {\left(x+2\right)}^2+{\left(y-1\right)}^2+z^2& =4+4+1& \hfill & \\ \hfill {\left(x+2\right)}^2+{\left(y-1\right)}^2+z^2& =3^2& \hfill & \end{array}$$

This is the equation of a sphere and the radius is $r=3$ and $s=\left(-2,1,0\right)$. is the center.