To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$\left(4x+1\right)^{2}=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $16$ for $a$, $8$ for $b$, and $-8$ for $c$ in the quadratic formula.
For the product to be $≤0$, one of the values $x-\frac{1}{2}$ and $x+1$ has to be $≥0$ and the other has to be $≤0$. Consider the case when $x-\frac{1}{2}\geq 0$ and $x+1\leq 0$.
$$x-\frac{1}{2}\geq 0$$ $$x+1\leq 0$$
This is false for any $x$.
$$x\in \emptyset$$
Consider the case when $x-\frac{1}{2}\leq 0$ and $x+1\geq 0$.
$$x+1\geq 0$$ $$x-\frac{1}{2}\leq 0$$
The solution satisfying both inequalities is $x\in \left[-1,\frac{1}{2}\right]$.
