#### Answer

There are 2 points $(-1,27)$ and $(2,0)$ for which the tangents there are parallel to the $x-$axis.

#### Work Step by Step

$$y=2x^3-3x^2-12x+20$$
1) First, find the derivative of $y$: $$y'=2\times3x^2-3\times2x-12+0$$ $$y'=6x^2-6x-12=6(x^2-x-2)$$
2) All the tangents being parallel to the $x-$axis, or in other words, horizontal tangents, have zero slopes.
This means we will see if there are any values of $x$ for which $y'=0$ or not; these values of $x$ represent points where the tangents are horizontal.
$$y'=0$$ $$6(x^2-x-2)=0$$ $$x^2-x-2=0$$ $$(x+1)(x-2)=0$$ $$x=-1\hspace{1cm}\text{or}\hspace{1cm}x=2$$
- For $x=-1$: $$y=2(-1)^3-3(-1)^2-12(-1)+20$$ $$=2(-1)-3(1)+12+20$$ $$=-2-3+32=27$$
- For $x=2$: $$y=2(2)^3-3(2)^2-12(2)+20$$ $$=2^4-3(4)-24+20$$ $$=16-12-4=0$$
Therefore, there are 2 points $(-1,27)$ and $(2,0)$ for which the tangents there are parallel to the $x-$axis.