Minimum Value of Quadratic Expression

Rewrite the expression by completing the square: $x^2 + 4xy + 6y^2 - 4y + 4 = (x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$ $= (x + 2y)^2 + \frac{1}{2}(4y^2 - 8y + 8) = (x + 2y)^2 + \frac{1}{2}[(2y - 2)^2 + 4]$ $= (x + 2y)^2 + \frac{1}{2}(2y - 2)^2 + 2$ Since squares of real numbers are non-negative, the minimum occurs when $x + 2y = 0$ and $2y - 2 = 0 \implies y = 1, x = -2$. Minimum value = $0 + 0 + 2 = 2$.