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x^{2}-5x\leq 0
Subtract 5x from both sides.
x\left(x-5\right)\leq 0
Factor out x.
x\geq 0 x-5\leq 0
For the product to be ≤0, one of the values x and x-5 has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-5\leq 0.
x\in \begin{bmatrix}0,5\end{bmatrix}
The solution satisfying both inequalities is x\in \left[0,5\right].
x-5\geq 0 x\leq 0
Consider the case when x\leq 0 and x-5\geq 0.
x\in \emptyset
This is false for any x.
x\in \begin{bmatrix}0,5\end{bmatrix}
The final solution is the union of the obtained solutions.