Multiply the inequality by -1 to make the coefficient of the highest power in $10x-x^{2}$ positive. Since $-1$ is negative, the inequality direction is changed.
$$-10x+x^{2}\leq 0$$
Factor out $x$.
$$x\left(x-10\right)\leq 0$$
For the product to be $≤0$, one of the values $x$ and $x-10$ has to be $≥0$ and the other has to be $≤0$. Consider the case when $x\geq 0$ and $x-10\leq 0$.
$$x\geq 0$$ $$x-10\leq 0$$
The solution satisfying both inequalities is $x\in \left[0,10\right]$.
$$x\in \begin{bmatrix}0,10\end{bmatrix}$$
Consider the case when $x\leq 0$ and $x-10\geq 0$.
$$x-10\geq 0$$ $$x\leq 0$$
This is false for any $x$.
$$x\in \emptyset$$
The final solution is the union of the obtained solutions.
