Show that the tangent to the curve y=(x^2+x-2)^3+3 at the point (1,3) is also the tangent to the curve at another point.

This is what I did so far...

I found the derivative of the curve and I got:

y'=6x^5+15x^4-16x^3-36x^2+16x+12

then I found

f'(1)=-3

and then I put

-3=6x^5+15x^4-16x^3-36x^2+16x+12

and now I am trying to solve for x, BUT I cannot find any factors of x... so? am I doing it right?