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 

since it has calculus tag im going to try solve it with calculus
heres my try:
we are going to find some extreme point and then check it if its minimum or maximum 
let $Z = 2x^2 +y^2 +2xy + 2x -3y +8 $
so we got $\frac{dz}{dx} =  4x + 2y +2 $ and $\frac{dz}{dx} =  2y +2x -3 $
now we are going to find a possible extreme point from the equation above
$ 4x + 2y + 2 = 0 $ change this to $ 4x+2y = -2$
$2x + 2y - 3 = 0 $ change this to  $ 2x + 2y = 3$
by eliminate 2 of the equation above we got x = $\frac{-5}{2}$ and y = $4$
and now we are going to check if this point is minimum or maximum .
by using second derivation $\frac{d^2z}{dx^2} = 4 $
now by subtitute $(x,y) = (\frac{-5}{2} ,4 )$ to the $\frac{d^2z}{dx^2} = 4 $
we got $\frac{d^2z}{dx^2} > 0 $, since $\frac{d^2z}{dx^2} > 0  $ this mean (x,y) 
is a extreme minimum . this mean by subtituting $(x,y) = (\frac{-5}{2} ,4 )$ we 
 are going to get a minimum result  
