Solve for x
x\in (-\infty,-\frac{9}{7}]\cup [0,\infty)
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-9x-7x^{2}\leq 0
Subtract 7x^{2} from both sides.
9x+7x^{2}\geq 0
Multiply the inequality by -1 to make the coefficient of the highest power in -9x-7x^{2} positive. Since -1 is negative, the inequality direction is changed.
x\left(7x+9\right)\geq 0
Factor out x.
x+\frac{9}{7}\leq 0 x\leq 0
For the product to be ≥0, x+\frac{9}{7} and x have to be both ≤0 or both ≥0. Consider the case when x+\frac{9}{7} and x are both ≤0.
x\leq -\frac{9}{7}
The solution satisfying both inequalities is x\leq -\frac{9}{7}.
x\geq 0 x+\frac{9}{7}\geq 0
Consider the case when x+\frac{9}{7} and x are both ≥0.
x\geq 0
The solution satisfying both inequalities is x\geq 0.
x\leq -\frac{9}{7}\text{; }x\geq 0
The final solution is the union of the obtained solutions.
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