Understanding how $x^2+2x+y^2-6y-15=0$ becomes $(x+1)^2+(y-3)^2=25$ 
I am looking at how this equation was simplified, $15$ was pushed to the other side and $10$ I believe was added by adding $1$ to the first bracket equation and $9$ to the other part.
The next part where to get $x² + 2x + 1$ to $(x + 1)^2$ there $2x$ seems to be dropped off and the same for $6y$.

Yes, you’re right. The equation was simplified by completing the square. Remember that $(m+n)^2 = m^2+2mn+n^2$ and vice versa.
$$x^2+\frac{b}{a}x = -c \implies x^2+\biggr(\frac{b}{2a}\biggr)^2+\frac{b}{a}x = -c+\biggr(\frac{b}{2a}\biggr)^2$$ 
$$\implies \biggr(x+\frac{b}{2a}\biggr)^2 = -c+\biggr(\frac{b}{2a}\biggr)^2 \implies \biggr(x+\frac{b}{2a}\biggr)^2-\biggr(\frac{b}{2a}\biggr)^2= -c$$
This process was done for both $x$ and $y$.
$$x^2+2x+y^2-6y-15 = 0$$
Break it down into two parts. Complete the square for both $x$ and $y$.
$$(x^2+2x)+(y^2-6y)-15 = 0$$
$$(x^2 + 1^2+2x)-1^2+(y^2+3^2-6y)-3^2-15 = 0$$
$$(x+1)^2-1+(y-3)^2-9-15 = 0 \implies (x+1)^2+(y-3)^2-25 = 0$$ $$\implies (x+1)^2+(y-3)^2 = 25$$
