Solve for x
x\in \begin{bmatrix}0,\frac{7}{2}\end{bmatrix}
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x\left(2x-7\right)\leq 0
Factor out x.
x\geq 0 x-\frac{7}{2}\leq 0
For the product to be ≤0, one of the values x and x-\frac{7}{2} has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-\frac{7}{2}\leq 0.
x\in \begin{bmatrix}0,\frac{7}{2}\end{bmatrix}
The solution satisfying both inequalities is x\in \left[0,\frac{7}{2}\right].
x-\frac{7}{2}\geq 0 x\leq 0
Consider the case when x\leq 0 and x-\frac{7}{2}\geq 0.
x\in \emptyset
This is false for any x.
x\in \begin{bmatrix}0,\frac{7}{2}\end{bmatrix}
The final solution is the union of the obtained solutions.
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