C sqrt [3](120+sqrt (28-sqrt {7+sqrt {4))}=

To solve this expression, we need to simplify the innermost square root first. <br /><br />The innermost square root is $\sqrt{4}$, which is equal to 2. <br /><br />Next, we have $\sqrt{7+\sqrt{4}}$, which simplifies to $\sqrt{7+2}$, or $\sqrt{9}$. <br /><br />The square root of 9 is 3. <br /><br />Now, we have $\sqrt{28-\sqrt{7+2}}$, which simplifies to $\sqrt{28-3}$, or $\sqrt{25}$. <br /><br />The square root of 25 is 5. <br /><br />Finally, we have $\sqrt[3]{120+\sqrt{28-\sqrt{7+2}}}$, which simplifies to $\sqrt[3]{120+5}$, or $\sqrt[3]{125}$. <br /><br />The cube root of 125 is 5. <br /><br />Therefore, the value of the given expression is 5.