Solve the quadratic equation $$ z^{2}-6 z+10+(z-6) i=0 $$ If $$ z_{1} $$ and $$ z_{2} $$ are the solutions, then $$ \left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}= $$

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To solve the quadratic equation

$$
z^{2} - 6z + 10 + (z - 6)i = 0,
$$

we can rearrange it as follows:

$$
z^{2} - 6z + 10 + zi - 6i = 0.
$$

This can be grouped into real and imaginary parts:

$$
z^{2} - 6z + (10 - 6i) + zi = 0.
$$

We can express $$ z $$ as $$ z = x + yi $$, where $$ x $$ and $$ y $$ are real numbers. Substituting $$ z $$ into the equation gives:

$$
(x + yi)^{2} - 6(x + yi) + (10 - 6i) + (x + yi - 6)i = 0.
$$

Expanding $$ (x + yi)^{2} $$:

$$
x^{2} + 2xyi - y^{2} = (x^{2} - y^{2}) + (2xy)i.
$$

Now substituting this into the equation:

$$
(x^{2} - y^{2}) + (2xy)i - 6x - 6yi + 10 - 6i + (x - 6 + yi)i = 0.
$$

This simplifies to:

$$
(x^{2} - y^{2} - 6x + 10) + (2xy - 6y + x - 6)i = 0.
$$

For the equation to hold, both the real and imaginary parts must equal zero:

1. Real part: 
   $$
   x^{2} - y^{2} - 6x + 10 = 0,
   $$
2. Imaginary part: 
   $$
   2xy - 6y + x - 6 = 0.
   $$

Now we will solve these equations. From the second equation, we can express $$ x $$:

$$
x = 6 - 2xy + 6y.
$$

Substituting $$ x $$ into the first equation:

$$
(6 - 2xy + 6y)^{2} - y^{2} - 6(6 - 2xy + 6y) + 10 = 0.
$$

This is quite complex, so let's instead use the quadratic formula directly on the original equation. We can rewrite the equation in standard form:

$$
z^{2} - 6z + (10 + (z - 6)i) = 0.
$$

This is a quadratic in $$ z $$:

$$
z^{2} - 6z + (10 - 6i) = 0.
$$

Using the quadratic formula $$ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:

Here, $$ a = 1 $$, $$ b = -6 $$, and $$ c = 10 - 6i $$.

Calculating the discriminant:

$$
b^2 - 4ac = (-6)^2 - 4(1)(10 - 6i) = 36 - 40 + 24i = -4 + 24i.
$$

Now we find $$ \sqrt{-4 + 24i} $$. To do this, we can express it in polar form. The modulus is:

$$
\sqrt{(-4)^2 + (24)^2} = \sqrt{16 + 576} = \sqrt{592} = 4\sqrt{37}.
$$

The argument $$ \theta $$ is given by:

$$
\theta = \tan^{-1}\left(\frac{24}{-4}\right) = \tan^{-1}(-6).
$$

Thus, we can express $$ -4 + 24i $$ in polar form as:

$$
4\sqrt{37} \text{cis}(\theta),
$$

where $$ \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) $$.

Now, we can find $$ \sqrt{-4 + 24i} $$:

$$
\sqrt{4\sqrt{37}} \text{cis}\left(\frac{\theta}{2}\right) = 2\sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right).
$$

Now substituting back into the quadratic formula:

$$
z = \frac{6 \pm 2\sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right)}{2} = 3 \pm \sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right).
$$

Let $$ z_1 = 3 + \sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right) $$ and $$ z_2 = 3 - \sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right) $$.

To find $$ |z_1|^2 + |z_2|^2 $$:

$$
|z_1|^2 = |3 + \sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right)|^2,
$$
$$
|z_2|^2 = |3 - \sqrt[4]{37} \text{cis}\left(\frac{\theta}{2}\right)|^2.
$$

Using the property $$ |a + b|^2 = |a|^2 + |b|^2 + 2\text{Re}(a\overline{b}) $$:

$$
|z_1|^2 + |z_2|^2 = 2|3|^2 + 2|\sqrt[4]{37}|^2 = 2(9) + 2(\sqrt{37}) = 18 + 2\sqrt{37}.
$$

Thus, the final answer is:

$$
\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2} = 18 + 2\sqrt{37}.
$$
$$ \left|z_{1}\right|^{2}+\left|z_{2}\right|^{2} = 18 + 2\sqrt{37}. $$

Solve the quadratic equation \[ z^{2}-6 z+10+(z-6) i=0 \] If \( z_{1} \) and \( z_{2} \) are the solutions, then \[ \left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}= \]

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