<s>[INST] How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$: 
$$ x^2 - 3x + 2y^2 + 4y + 2. $$
I've been thinking about maybe somehow making $c$ equal $2y^2 + 4y + 2$?
The answer from the answerbook is: $$ x^2 - 3x + 2y^2 + 4y + 2 = \left( x - \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 - \dfrac {9}{4}. $$
 [/INST] EDIT:
$$ x^2 - 3x + 2y^2 + 4y + 2$$ 
can be re-written as
$$ x^2-2*x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
Obeserve that the first two terms in the expression are of the form 
$$a^2+2a.b+b^2=(a+b)^2$$
Hence we get
$$ x^2 - 3x + 2y^2 + 4y + 2=x^2-2x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
$$\Rightarrow x^2 - 3x + 2y^2 + 4y + 2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$$ 
On the RHS we have the sum of two squares is is . The minimum value of each square term will be 0 as a square term cannot be negative in $IR$. Hence the minimum value of the given expression is -$\frac{9}{4}$ when the two square terms are 0(at x=$\frac{3}{2}$ and y=-1).
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