$\sqrt{19+4\sqrt{12}}+\sqrt{19-4\sqrt{12}}$

Note that:

$19 + 4\sqrt{12} = 19 + 8\sqrt{3} = 16 + 3 + 2(4\sqrt{3})$

$= 4^2 + (\sqrt{3})^2 + 2(4\sqrt{3}) = 4^2 + 2(4\sqrt{3}) + (\sqrt{3})^2$

Now,

$4^2 + 2(4\sqrt{3}) + (\sqrt{3})^2$ is equivalent to $(4 + \sqrt{3})^2$ . That is

$19 + 4\sqrt{12}$ = $(4 + \sqrt{3})^2$ . Taking the square root we have $4 + \sqrt{3}$

Similarly, $19 - 4\sqrt{12}$ = $(4 - \sqrt{3})^2$ . Taking the square root we have $4 - \sqrt{3}$ .

Thus,

$\sqrt{19 + 4\sqrt{12}} + \sqrt{19 - 4\sqrt{12}} = (4 + \sqrt{3}) + (4 - \sqrt{3}) = 8$

let the given expression be equal to x.......... squaring both sides.............and then simplifying ......x=8

by calculator....:D