Trigonometric solution for finding the lengths of an obtuse triangle given the base, angle, & area

Given a triangle ABC where $$\measuredangle BAC = 120^\circ$$, $$BC = \sqrt{37}$$, and area $$3\sqrt{3}$$, find the lengths AB and AC.

Here is my attempt:

At first I thought of doing it by getting BA by $$BA = BP - AP$$ then use the two lengths and angle formula for triangle area having $$\sin \alpha$$ as my variable, but the problem is that I cannot define BP in terms of $$sin\alpha$$, and get an equation with $$\cos \alpha$$ instead where they have different factors that prevent me from applying any trigonometric identity.
I've also tried doing $$3\sqrt{3} = \frac{PC \cdot BA}{2}$$, but I run into the same problem.

Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.

Guide:

Let $$AB=y$$,

By cosine rule $$x^2+y^2-2xy \cos 120^\circ = 37$$

From the area, we know that

$$\frac12 xy \sin 120^\circ = 3\sqrt3$$

$$xy=12\tag{1}$$

and $$x^2+y^2+12=37$$

$$x^2+y^2=25=5^2\tag{2}$$

Can you solve for $$x$$ and $$y$$ from $$(1)$$ and $$(2)$$? (Further hint: I purpose write $$25=5^2$$, if you are familiar with certain identities, the answer could be very obvious).