Let's solve the given equations step by step.

### Equation 1: \( 2x \cdot x + 6 + 5x + 6 = 0 \)

First, simplify the equation:

\[ 2x^2 + 5x + 6 + 6 = 0 \]

Combine like terms:

\[ 2x^2 + 5x + 12 = 0 \]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = 5 \), and \( c = 12 \).

To solve this quadratic equation, we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute \( a = 2 \), \( b = 5 \), and \( c = 12 \):

\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 12}}{2 \cdot 2} \]
\[ x = \frac{-5 \pm \sqrt{25 - 96}}{4} \]
\[ x = \frac{-5 \pm \sqrt{-71}}{4} \]

Since the discriminant (\( \Delta = b^2 - 4ac \)) is negative (\( -71 \)), the equation has no real solutions. The solutions are complex numbers:

\[ x = \frac{-5 \pm i\sqrt{71}}{4} \]

### Equation 2: \( x \cdot 3x + 5 - 6x - 10 = 0 \)

First, simplify the equation:

\[ 3x^2 + 5 - 6x - 10 = 0 \]

Combine like terms:

\[ 3x^2 - 6x - 5 = 0 \]

This is another quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -6 \), and \( c = -5 \).

Again, use the quadratic formula to solve for \( x \):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute \( a = 3 \), \( b = -6 \), and \( c = -5 \):

\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
\[ x = \frac{6 \pm \sqrt{36 + 60}}{6} \]
\[ x = \frac{6 \pm \sqrt{96}}{6} \]
\[ x = \frac{6 \pm 4\sqrt{6}}{6} \]
\[ x = 1 \pm \frac{2\sqrt{6}}{3} \]

So, the solutions are:

\[ x = 1 + \frac{2\sqrt{6}}{3} \]
\[ x = 1 - \frac{2\sqrt{6}}{3} \]

In summary:

- The first equation \( 2x^2 + 5x + 12 = 0 \) has no real solutions, but has complex solutions \( x = \frac{-5 \pm i\sqrt{71}}{4} \).
- The second equation \( 3x^2 - 6x - 5 = 0 \) has real solutions \( x = 1 + \frac{2\sqrt{6}}{3} \) and \( x = 1 - \frac{2\sqrt{6}}{3} \).