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A school is holding a play. They have calculated that if they charge $\$10$ for the play, then $300$ people will attend the play. They have also calculated that for each $\$0.25$ decrease for the admission fee, $10$ more people will attend the play. How much should the school charge for admission to maximize profit?

My equation: $$(10-0.25x)(300+10x)=P$$ Expanding I get $$-2.5x^2+25x+3000=P$$ Using the formula $-\dfrac b{2a}$, I get that $x=\dfrac{-25}{-2.5}=10$. Then, plugging this back into the equation, I get that the school should charge $\$7.50$ for the play. My answer was incorrect; the correct answer is $\$8.75$. What did I do wrong?

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    You forgot the $2$ in the denominator: what you calculated for $x$ was $-\frac{b}{a}$. – 2017-01-19
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    See the very similar [Convert this word problem into a system of equations](http://math.stackexchange.com/questions/2095730/convert-this-word-problem-into-a-system-of-equations). – 2017-01-19

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yus another quadratic

Anyway, your $-\frac{b}{2a}$ is wrong this time; it should give $x=\frac{-25}{2(-2.5)}=5.$ So $F=8.75$ as a result, which is what we want.

(F is the fee)