Minimum value of the given function 
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .

Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$ 
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?

By letting $x=t+1$ we get an even function
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}.$$
Now we show that the minimum value is attained at $t=0$: we have to verify
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}\geq \sqrt{20}$$
or, after squaring,
$$4t^2+10+2\sqrt{4t^4-16t^2+25}\geq 20$$
that is
$$\sqrt{(5-2t^2)^2+4t^2}\geq 5-2t^2$$
which trivially holds.
