Step 1: Frame the problem using a positive variable.
Since we already know both integers are negative, let's name them in a way that keeps our working variable positive throughout, instead of solving for a negative $x$ and then checking its sign at the end. Let the smaller (more negative) number be $-(n+1)$ and the larger number be $-n$, where $n$ is a positive whole number. For any positive $n$, these two are automatically consecutive negative integers.
Step 2: Write the given condition as an equation in n.
The sum of their squares is $481$, so:
\[ \left(-(n+1)\right)^2 + (-n)^2 = 481 \]
Squaring removes the negative sign, so this is the same as:
\[ (n+1)^2 + n^2 = 481 \]
Step 3: Expand and simplify.
\[ n^2 + 2n + 1 + n^2 = 481 \]
\[ 2n^2 + 2n + 1 = 481 \]
Subtract $481$ from both sides:
\[ 2n^2 + 2n - 480 = 0 \]
Divide the whole equation by $2$ to simplify:
\[ n^2 + n - 240 = 0 \]
Step 4: Solve this quadratic using the quadratic formula.
Here $a = 1$, $b = 1$, $c = -240$.
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 960}}{2} = \frac{-1 \pm \sqrt{961}}{2} = \frac{-1 \pm 31}{2} \]
This gives two values:
\[ n = \frac{30}{2} = 15 \quad \text{or} \quad n = \frac{-32}{2} = -16 \]
Step 5: Pick the valid value of n and write down the two integers.
By our own setup in Step 1, $n$ was defined to be a positive whole number, so we reject $n = -16$ and keep $n = 15$.
The two consecutive negative integers are then:
\[ -n = -15 \quad \text{and} \quad -(n+1) = -16 \]
Final Answer:
The two consecutive negative integers are:
\[ \boxed{-16 \text{ and } -15} \]