Let the area enclosed by the $x$-axis, and the tangent and normal drawn to the curve $4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$  at the point $\left(-2,3\right)$ be $A$. Then $8A$ is equal to _______.

The given curve is $4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ 

On differentiating both sides, we get

$12x^2-3y^2-6xy{y}^{\text{'}}+12x-5y-5x{y}^{\text{'}}-16y{y}^{\text{'}}+9=0$

Now, at the point $\left(-2,3\right)$

$48-27+36y^{\text{'}}-24-15+10y^{\text{'}}-48y^{\text{'}}+9=0$

i.e. ${\left(y^{\text{'}}\right)}_{\left(-2,3\right)}=\frac{-9}{2}$

So slope of tangent $m_T=\frac{-9}{2}$ and 

slope of normal $m_N=\frac{2}{9}$

i.e. equation of tangent $T\equiv y-3=\frac{-9}{2}\left(x+2\right)\Rightarrow y=-\frac{9}{2}x-6$

and equation of normal $N\equiv y-3=\frac{2}{9}\left(x+2\right)\Rightarrow y=\frac{2}{9}x+\frac{31}{9}$

Intersection point of tangent and normal is $\left(-2,3\right)$

The intersection point of tangent and normal with $x$-axis respectively will be $\left(-\frac{4}{3},0\right) \& \left(-\frac{31}{2},0\right)$

Now, area of triangle formed $=\frac{1}{2}\times$Base$\times$Height

$A=\frac{1}{2}\times \left(\frac{-4}{3}+\frac{31}{2}\right)\left(3\right)=\frac{1}{2}\left(\frac{85}{6}\right)·3=\frac{85}{4}$

$\Rightarrow 8A=170$