What is the method to solve $\min(5-15a-18b+15a^2+45ab+45b^2)$ 
Let $(a,b)\in \mathbb{R}$
$A:=5-15a-18b+15a^2+45ab+45b^2$

My question is how to find $\min(A)$
I know the method is to rearrange this expression in
$A=(A_1 )^2+ (A_2)^2+\alpha$
My question is :is there a method?
 A: You can complete the squares.  First write 
$$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+\frac {45}4(a+2b)^2+\frac {15}4a^2$$
where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient.  That got rid of the cross term.  Now we can write 
$$5-15a-18b+\frac {45}4(a+2b)^2+\frac {15}4a^2=\frac {45}4(a+2b)^2-9(a+2b)+\frac {15}4a^2-6a+5$$
and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is.  Then set $a=-d, b=\frac 12(-c-a)$ to zero out the squares and you are left with the constant.
A: Ok here are two methods. 
The first and imo easiest is to consider this as two single variable problems. Consider 
$$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=\frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=\frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=\frac{-(By+D)}{2A}$,$y=\frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$
The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum.  To do this you rotate the $x$ and $y$ axis around by some angle $\theta$ to get a new $X$ and $Y$ axis.   

Now it turns out that $x=Xcos\theta-Ysin\theta$ and $y=Xsin\theta + Ycos\theta$. And it also turns out that there will be no $XY$ term if $cot 2\theta =\frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
 and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.
