Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0 at the point (–2, 3) be A. Then 8A is equal to _____

$$\begin{gathered} 4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0\text{ differentiating both sides we get } \\ 12x^2-3y^2-6xy\frac{dy}{dx}+12x-5y-5x\frac{{\displaystyle dy}}{{\displaystyle dx}}-16y\frac{{\displaystyle dy}}{{\displaystyle dx}}+9=0 \\ Substituting \left(-2,3\right) \\ \begin{array}{l}\Rightarrow 48-27+36\frac{dy}{dx}-24-15+10\frac{dy}{dx}-48\frac{dy}{dx}+9=0\\ \\ \Rightarrow \frac{dy}{dx}=\frac{-9}{2}\\ m=slope of \tan gent=\frac{-9}{2}\\ \end{array} \end{gathered}$$the area enclosed by the x-axis, and the tangent and normal =$$\begin{gathered} A=\frac{1}{2}\left|\left(Length of \tan gent \right)\left(length of normal\right)\right| \\ \Rightarrow A=\frac{1}{2}\left|\left(\frac{y}{m}\sqrt{1+m^2}\right)\left(y\sqrt{1+m^2}\right)\right| \\ \Rightarrow A=\left|\frac{y^2\left(1+m^2\right)}{2m}\right|=\frac{9\left(1+{\displaystyle \frac{81}{4}}\right)}{2\left({\displaystyle \frac{-9}{2}}\right)}=\frac{85}{4} \\ \Rightarrow 8A=170 \end{gathered}$$