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Solution
$$2x^2+x-528=0$$
$$2x^2+33x-32x-528=0$$
$$x(2x+33)-16(2x+33)=0$$
$$(2x+33)(x-16)=0$$
$$2x+33=0$$ or $$x-16=0$$
Therefore, $$x=\dfrac{-33}{2}$$ or $$16$$.
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