Finding the two points on the curve $y=x^3 + 3x + 4$ at which the gradient is $6$ I already was able to find out the first coordinate, which is $(1,8)$; however I can't seem to find the second one.

Any sort of question that involves finding the gradient at a certain point should give you the signal to use differentiation.
Differentiating our function with respect to $x$ then gives $$\frac{dy}{dx}=3x^2+3.$$ Now, they've given you the gradient, namely $6$, so we set our derivative equal to this and we get $$3x^2+3=6\Rightarrow x=\pm 1.$$ Finally, we must substitute our values of $x$ into our original function to obtain the $y$ values. So, we get $$y=1^3+3(1)+4=8,\quad y=(-1)^3+3(-1)+4=0.$$ So the points are $(1,8)$ and $(-1,0)$.
