Let the function be
\[
f(x)=2x^3+3x^2-36x+10.
\]
To find the point where the function has a minimum, first find the critical points by differentiating.
\[
f'(x)=\frac{d}{dx}(2x^3+3x^2-36x+10).
\]
\[
f'(x)=6x^2+6x-36.
\]
Take \(6\) common:
\[
f'(x)=6(x^2+x-6).
\]
Factorize:
\[
x^2+x-6=(x+3)(x-2).
\]
So,
\[
f'(x)=6(x+3)(x-2).
\]
For critical points:
\[
f'(x)=0.
\]
\[
6(x+3)(x-2)=0.
\]
Therefore,
\[
x=-3
\]
or
\[
x=2.
\]
Now use the second derivative test.
\[
f''(x)=\frac{d}{dx}(6x^2+6x-36).
\]
\[
f''(x)=12x+6.
\]
At \(x=2\):
\[
f''(2)=12(2)+6=24+6=30.
\]
Since
\[
f''(2)>0,
\]
the function has a minimum at
\[
x=2.
\]
Hence, the required value of \(x\) is
\[
2.
\]