$x \times \left( 5x+7 \right) \geq 0$
Separate the inequality into two possible cases$\begin{array} { l }\left\{\begin{array} { l } x \geq 0 \\ 5x+7 \geq 0\end{array} \right.,\\\left\{\begin{array} { l } x \leq 0 \\ 5x+7 \leq 0\end{array} \right.\end{array}$
Solve the inequality for $x$$\begin{array} { l }\left\{\begin{array} { l } x \geq 0 \\ x \geq -\frac{ 7 }{ 5 }\end{array} \right.,\\\left\{\begin{array} { l } x \leq 0 \\ 5x+7 \leq 0\end{array} \right.\end{array}$
Solve the inequality for $x$$\begin{array} { l }\left\{\begin{array} { l } x \geq 0 \\ x \geq -\frac{ 7 }{ 5 }\end{array} \right.,\\\left\{\begin{array} { l } x \leq 0 \\ x \leq -\frac{ 7 }{ 5 }\end{array} \right.\end{array}$
Find the intersection$\begin{array} { l }x \in \left[ 0, +\infty\right\rangle,\\\left\{\begin{array} { l } x \leq 0 \\ x \leq -\frac{ 7 }{ 5 }\end{array} \right.\end{array}$
Find the intersection$\begin{array} { l }x \in \left[ 0, +\infty\right\rangle,\\x \in \left\langle-\infty, -\frac{ 7 }{ 5 }\right]\end{array}$
Find the union$x \in \left\langle-\infty, -\frac{ 7 }{ 5 }\right] \cup \left[ 0, +\infty\right\rangle$