Real ordered pair $(a,b)$ in equation 
If $$a^2+5b^2+2b=6a+2ab-10$$
then all  real ordered pair of $(a,b)$ is?

Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.

$a^2+5b^2+2b-6a-2ab+10=0$
Let's suppose we had a $(a+kb+j)^2$ term and that were the only term with $a $.  We'd have $(a+kb+j)^2=a^2+k^2b^2+j^2+2kab+2aj+2kjb $.  So $k=-1;j=-3$.
So
$a^2+5b^2+2b-6a-2ab+10=0$
$a^2+b^2+9-2ab-6a+6b+4b^2-4b+1=0$
$(a-b-3)^2+(2b-1)^2=0$
So for real solutions 
$2b-1=0$ and $b=\frac 12$
And $a-b-3=0$ so $a=3\frac 12$.
