O Radicals and Quadratic Functions Solving a radical equation with two radicals that simplifies to sqrt $$ (\mathrm{x})=\mathrm{a} $$ Solve for $$ y $$, where $$ y $$ is a real number. $$ 5+\sqrt{y}=\sqrt{y+55} $$ If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

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To solve the equation:

$$
5 + \sqrt{y} = \sqrt{y + 55}
$$

**Step 1: Isolate the radicals**

First, let's isolate one of the square roots:

$$
\sqrt{y + 55} - \sqrt{y} = 5
$$

**Step 2: Substitute and simplify**

Let $$ \sqrt{y} = a $$. Then, $$ \sqrt{y + 55} = a + 5 $$.

Substitute into the equation:

$$
(a + 5)^2 = a^2 + 55
$$

Expand and simplify:

$$
a^2 + 10a + 25 = a^2 + 55
$$

$$
10a + 25 = 55
$$

$$
10a = 30
$$

$$
a = 3
$$

**Step 3: Solve for $$ y $$**

Since $$ a = \sqrt{y} $$:

$$
\sqrt{y} = 3 \implies y = 9
$$

**Step 4: Verify the solution**

Substitute $$ y = 9 $$ back into the original equation:

$$
5 + \sqrt{9} = 5 + 3 = 8
$$
$$
\sqrt{9 + 55} = \sqrt{64} = 8
$$

Both sides are equal, so $$ y = 9 $$ is a valid solution.

**Conclusion**

The solution to the equation is:

$$
\boxed{9}
$$
$$ y = 9 $$

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