Find the y-coordinate of the point.

Substitute $x=2$ into the equation of the curve $y = 2x^2 - 3x - 6$.

$$y = 2(2)^2 - 3(2) - 6$$ $$y = 2(4) - 6 - 6$$ $$y = 8 - 6 - 6$$ $$y = 2 - 6$$ $$y = -4$$

The point is $(2, -4)$.

Differentiate $y = 2x^2 - 3x - 6$ with respect to $x$.

$$\frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) - \frac{d}{dx}(6)$$ $$\frac{dy}{dx} = 4x - 3$$

Substitute $x=2$ into the derivative $\frac{dy}{dx}$.

$$m_t = 4(2) - 3$$ $$m_t = 8 - 3$$ $$m_t = 5$$

The slope of the normal $m_n$ is the negative reciprocal of the slope of the tangent $m_t$.

$$m_n = -\frac{1}{m_t}$$ $$m_n = -\frac{1}{5}$$

Using the point $(x_1, y_1) = (2, -4)$ and the slope $m_n = -\frac{1}{5}$:

$$y - y_1 = m_n(x - x_1)$$ $$y - (-4) = -\frac{1}{5}(x - 2)$$ $$y + 4 = -\frac{1}{5}(x - 2)$$

Multiply both sides by 5 to clear the fraction:

$$5(y + 4) = -1(x - 2)$$ $$5y + 20 = -x + 2$$

Rearrange the equation to the standard form $Ax + By + C = 0$:

$$x + 5y + 20 - 2 = 0$$ $$x + 5y + 18 = 0$$