Answer
$x \leq -2$
Work Step by Step
$x^{2} + 4x\leq x(x-5)-18$
$x^{2} + 4x \leq x^{2} - 5x - 18$
$x^{2} - x^{2} + 4x + 5x + 18 \leq 0$
$9x + 18 \leq 0$
$9x \leq -18$
$x \leq -2$
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