Finding the exact value of a radical

If you suspect that $\sqrt{97+56\sqrt3}$ (or similar) equals something like $a+\sqrt b$ with $a$, $b$ rational you must have $$97+56\sqrt 3=a^2+b+2a\sqrt b.$$ To match up the surds you'll have to have $a^2+b=97$ and $2a\sqrt b=56\sqrt3$. Thus $a^2b=28^2\times 3=2352$. Therefore $a^2$ and $b$ are the solutions of the quadratic $$x^2-97x+2352=0.$$ The solutions are $x=48$ and $x=49$. Then $a^2=49$ as that's the one that's a square, and $b^2=48$. So $$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$

Wikipedia describes a simple method to write $$ \sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e} $$ with $$ d=\frac{a + \sqrt {a^2-b^2c}}{2}, \qquad e=\frac{a - \sqrt {a^2-b^2c}}{2} $$ This works iff $a^2 - b^2c$ is a square.

For $\sqrt{97 +56\sqrt3}$ we have $a^2 - b^2c=1$ and so $$ \sqrt{97 +56\sqrt3} = \sqrt{49}+\sqrt{48} = 7 + 4 \sqrt 3 $$

$$\sqrt{97 +56\sqrt3}=\sqrt{49+48 +56\sqrt3} =\sqrt{7^2+(4\sqrt 3)^2 +2 \times 7 \times 4\sqrt3} = \sqrt{(7+4\sqrt3)^2 }= 7 +4\sqrt3 $$