least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$ 
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$

$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
So $$(2x-3)^2-4(2x^2+2x+8-K)\geq 0$$
So $$-4x^2-20x-23+4K\geq0\Rightarrow 4x^2+20x+23-4K\leq0$$
Now How can I Solve after that, Help me
Thanks 

You have
$$4x^2+20x+23-4K\le 0.$$
So, 
$$\begin{align}K&\ge x^2+5x+\frac{23}{4}\\&=\left(x+\frac 52\right)^2-\frac 12\\&\ge -\frac 12\end{align}$$
The equality is attained when $x=-\frac 52,y=4$.
