Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$. 
Then i calculated the area with 
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next. 

If $AB=x$, then $AC=x-2,BC=x+1$.
Now use the cosine formula: $$x^2=(x-2)^2+(x+1)^2-(x-2)(x+1)$$ Expanding we get $x=7$. So the longest side is 8 and the height $AC\sin60^o=5\sqrt3/2$. Hence the area is $10\sqrt3$.
