<s>[INST] 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.
 [/INST] The equation can be written as 
$$(2b-1)^2+ a^2+b^2+9-2ab-6a+6b=0$$
$$\Rightarrow (2b-1)^2+ (-a)^2+(b)^2+(3)^2+ (2)(-a)(b)+(2)(3)(-a) +(2)(3)(b)=0$$
$$\Rightarrow (2b-1)^2+(a-b-3)^2=0$$
Hence we get $(a, b) = (\frac {7}{2} , \frac {1}{2})$
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