Prove that a degree-$6$ polynomial has exactly $2$ real roots

Hint: by Descartes' rule of signs the equation has no positive real roots, and at most $2$ negative ones. But you showed that it has at least one real root (and it's enough that $\,f(-1/2) \lt 0 \lt f(0)\,$ for that), then it must have a second real one, since non-real complex roots come in conjugate pairs.

$$f''(x)=7\cdot6\cdot5x^4\geq0,$$ which says that $f$ is a convex function.

Thus, $f$ has two roots maximum and by your work we get two roots exactly.

Hint

You proved so far that there are at least 2 solutions.

If the function would have 3 or more solutions, then by Rolle's Theorem, $f'$ would have at least 2 solutions.