# Math Help - Quadratic equation

When bicycles are sold for $300 each, a cycle store can sell 70 in a season. For every$25 increase in the price, the number sold drops by 10.
a) Represent the sales revenue as a function of the price.
b)The total sales revenue is $17,500. How many bicycles were sold? What is the price of one bicycle? c) What range of prices will give a sales revenue that exceeds$18,000?
I'm stuck on b
R= revenue
n= number of bikes
p= price
For a
Price| Number of Bikes sold
300 | 70
325 | 60
350 | 50
So find the slope
$\frac{(70-50)}{(300-350)}=-0.4$
Plug that into y=mx+b
$70=-0.4(300)+b$
$70=-120+b$
$70+120=190$
$b=190$
So that means the function to find n(number of bicycles sold) is $-0.4p+190$
Since Revenue is $r=n*p$ then part a is $r=p(-0.4p+190)$ which simplifies to $r=-0.4p^2+190p$
So to solve part b?

$17500=-0.4p^2+190p$

2. I got it
$17500=-0.4p^2+190p$
$\frac{17500}{-0.4}=-43750$
$-43750=p^2+(\frac{190}{-0.4}p$
$0=p^2-475p+43750$
$0=(p-125)(p-350)$

$p=125, 350$