Find the number if eight times of its is added to its square, the sum so obtained is -16.

Answer

To solve the problem, we need to find a number \( x \) such that when eight times the number is added to its square, the result is -16. Let's break this down step by step.

### Step 1: Set up the equation
We start by translating the problem into a mathematical equation. The problem states that:

\[ x^2 + 8x = -16 \]

### Step 2: Rearrange the equation
To solve for \( x \), we need to rearrange the equation so that one side equals zero. We do this by adding 16 to both sides:

\[ x^2 + 8x + 16 = 0 \]

### Step 3: Factor the quadratic equation
Next, we will factor the quadratic equation \( x^2 + 8x + 16 = 0 \). We look for two numbers that multiply to 16 (the constant term) and add up to 8 (the coefficient of \( x \)). The numbers 4 and 4 fit this requirement:

\[ (x + 4)(x + 4) = 0 \]

This can be written as:

\[ (x + 4)^2 = 0 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \) by setting the factor equal to zero:

\[ x + 4 = 0 \]

Subtracting 4 from both sides gives:

\[ x = -4 \]

### Conclusion
Thus, the number we are looking for is:

\[ \boxed{-4} \]