Given equation of curve is

$y=2x^3-15x^2+36x-8 ...\left(1\right)$

 To find the turning points of the curve

$y^{\text{'}}=0$

$\frac{d}{dx}\left(2x^3-15x^2+36x-8\right)=0$

$6x^2-30x+36=0$

$x^2-5x+6=0$

$x^2-3x-2x+6=0$

$\left(x-3\right)\left(x-2\right)=0$

$\therefore x=3$ or $x=2$

Substitute $x=3$ in equation $\left(1\right)$

$y=2{\left(3\right)}^3-15{\left(3\right)}^2+36\left(3\right)-8=19$

Substitute $x=2$ in equation $\left(1\right)$

$y=2{\left(2\right)}^3-15{\left(2\right)}^2+36\left(2\right)-8=20$

$\therefore$ The turning points are $\left(3,19\right)$ and $\left(2,20\right)$

Here, 

$\left(\alpha ,\beta \right)\equiv \left(2,20\right) \& \left(y,\delta \right)\equiv \left(3,19\right) \left\{\because \alpha <y\right\}$

$\Rightarrow \alpha -y-\beta +\delta =2-3-20+19=-2$