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.

Consider the equation as a quadratic in $\,a\,$:
$$
a^2 - 2a(b+3) +5b^2+2b + 10 = 0
$$
Then its reduced discriminant is:
$$
\frac{1}{4} \Delta = (b+3)^2-(5b^2+2b +10) = -4b^2 + 4b -1 = -(2b-1)^2 \;\;\le\;\; 0
$$
It follows that $\,\Delta = 0\,$ for real roots to exist, so $\,b=\dfrac{1}{2}\,$, then substituting back $\,a=b+3=\dfrac{7}{2}\,$.
