Multiply both the numerator and denominator by [tex]2+\sqrt3[/tex], which is called the "conjugate" of [tex]2-\sqrt3[/tex]:
[tex]\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}[/tex]
Why do we pick this number? Recall the difference of squares factorization:
[tex]a^2-b^2=(a-b)(a+b)[/tex]
Now replace [tex]a=2[/tex] and [tex]b=\sqrt3[/tex]. Then
[tex](2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1[/tex]
So in your fraction, you end up with
[tex]\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)[/tex]
Finally, just expand the product.
[tex](5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}[/tex]
