If you know about the Intermediate value theorem, you can use the fact that you have $f(x)=6x^4-7x+1$, and that this function is continuous. Consider $f(-1)=6+7+1$, we have that $f(-1)>0$. now consider $f\Big(\frac{1}{2}\Big)=-\frac{17}{8}<0$. So the IVT say that for $f(a)\neq{f(b)}$, we have that $\forall\;y$ between $f(a)$ and $f(b)$, $\exists\;c\in\;]a,b[$ such as $f(c)=y$. We have that 0 is between $f(-1)$ and $f\Big(\frac{1}{2}\Big)$ so $\exists\;c\in]-1,\frac{1}{2}[$ such as $f(c)=0$. You can do the same for the fact that $0$ is between $f(2)$ and $f\Big(\frac{1}{2}\Big)$ so $\exists\;d\in\;]\frac{1}{2},2[$ such as $f(d)=0$.
We proved that there is two distinct roots. Now we must prove that there is only two roots. We can see if $f(x)$ is increasing for $x>2$.
\begin{equation}
f(x+1)-f(x)=6(x+1)^4-7(x+1)+1-6x^4+7x-1\\
=6x^4+24x^3+36x^2+24x+6-7x-7-6x^4+7x\\
=24x^3+36x^2+24x-1>0,\;\forall\;x>0
\end{equation}
So $f(x)$ is always increasing.
So if we consider $M\in\mathbb{R}$ such as $M=f(x),\;\forall\;x\in]2,+\infty[$. We have that $0$ can't be between $f(2)$ and $f(M)$ and by the IVT, it's impossible to find $f(e)=0$ such as $e\in\;]2,M[$.
You can use the same strategy to prove that there is no other roots in the negative numbers.

This is your function.