Find the value of $c$ such that $h(k(x))=0$ has equal roots, where $h(x)=5x+2$ and $k(x)=cx^2-x+2$ First I rearranged the equation:
$$h(k(x))=5(cx^2-x+2)+2=0$$
$$5cx^2-5x+12=0$$
So the next step would be to set the discriminant equal to $0$:
$$25-240c=0$$
$$c=\frac{5}{48}$$
Would $\frac{5}{48}$  be the value of $c$?

As the commenters confirmed, you're correct that $c=\frac{5}{48}$. The resulting quadratic equation will have one root of multiplicity two, which is $x=4.8$.
