$$(9x-7)^{2}-10\le(9x+3)(9x-5)$$
$x\geq \frac{1}{2}$
$$81x^{2}-126x+49-10\leq \left(9x+3\right)\left(9x-5\right)$$
$$81x^{2}-126x+39\leq \left(9x+3\right)\left(9x-5\right)$$
$$81x^{2}-126x+39\leq 81x^{2}-18x-15$$
$$81x^{2}-126x+39-81x^{2}\leq -18x-15$$
$$-126x+39\leq -18x-15$$
$$-126x+39+18x\leq -15$$
$$-108x+39\leq -15$$
$$-108x\leq -15-39$$
$$-108x\leq -54$$
$$x\geq \frac{-54}{-108}$$
$$x\geq \frac{1}{2}$$
Show Solution
Hide Solution
