13. $$ x=-6 \pm i \sqrt{6} $$ with $$ y $$-intercept $$ (0,30) $$

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To solve the problem, we have a complex number $$ x = -6 \pm i \sqrt{6} $$ and a $$ y $$-intercept at the point $$ (0, 30) $$.

1. **Extract Known Conditions:**
   - The real part of $$ x $$ is $$-6$$.
   - The imaginary part of $$ x $$ is $$ \pm i \sqrt{6} $$.
   - The $$ y $$-intercept is at the point $$ (0, 30) $$.

2. **Understanding the Context:**
   - The expression for $$ x $$ suggests that it could be part of a quadratic equation, where the roots are complex.
   - The $$ y $$-intercept indicates that when $$ x = 0 $$, $$ y = 30 $$.

3. **Finding the Quadratic Equation:**
   - The roots of a quadratic equation can be expressed as $$ x = r_1 $$ and $$ x = r_2 $$. Here, $$ r_1 = -6 + i \sqrt{6} $$ and $$ r_2 = -6 - i \sqrt{6} $$.
   - The sum of the roots $$ r_1 + r_2 = -12 $$ and the product of the roots $$ r_1 \cdot r_2 = (-6)^2 + (\sqrt{6})^2 = 36 + 6 = 42 $$.

4. **Forming the Quadratic Equation:**
   - The quadratic equation can be formed using the roots:
   $$
   y = a(x^2 + 12x + 42)
   $$
   - To find $$ a $$, we can use the $$ y $$-intercept. When $$ x = 0 $$, $$ y = 30 $$:
   $$
   30 = a(0^2 + 12 \cdot 0 + 42) \implies 30 = 42a \implies a = \frac{30}{42} = \frac{5}{7}
   $$

5. **Final Quadratic Equation:**
   - Substituting $$ a $$ back into the equation gives:
   $$
   y = \frac{5}{7}(x^2 + 12x + 42)
   $$

Now, let's summarize the final quadratic equation:

$$
y = \frac{5}{7}(x^2 + 12x + 42)
$$

This equation represents the relationship between $$ x $$ and $$ y $$ given the conditions provided.
The quadratic equation is $$ y = \frac{5}{7}(x^2 + 12x + 42) $$.

13. \( x=-6 \pm i \sqrt{6} \) with \( y \)-intercept \( (0,30) \)

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