Answer
$(-1,27)$ and $(2,0)$
Work Step by Step
Step 1. Find the derivative of the function $y=2x^3-3x^2-12x+20$ as $y'=6x^2-6x-12$
Step 2. For the tangent to be parallel to the x-axis, let $y'=0$; we have $6x^2-6x-12=0$ which gives $x^2-x-2=0$ or $(x+1)(x-2)=0$. Thus, we get $x_1=-1, x_2=2$
Step 3. We see that
$y_1=2(-1)^3-3(-1)^2-12(-1)+20=27$ and $y_2=2(2)^3-3(2)^2-12(2)+20=0$
Step 4. Thus the answer is yes and the points are $(-1,27)$ and $(2,0)$
