Cost and revenue - break even points

[kickingtoad]

The total revenue and total cost functions for the production and sale of x TV's are given as

$\displaystyle {R(x)=190x-0.2x^{2}}$
$\displaystyle {C(x)=3550+24x}$

List the values of x at the break even point(s). It is possible that there are no break even points.

I think to get the break even points you subtract C(x) from R(x) and set it equal to 0.

$\displaystyle {R(x)-C(x)=0}$

$\displaystyle {190x-0.2x^{2}-(3550+24x)=0}$

$\displaystyle {-0.2x^{2}+166x-3550=0}$

Where do I go from here?

 

[galactus]

This is not a calculus problem. It is being moved to Intermediate Algebra.

Use the quadratic formula to solve for x. There will be two values.

 

[kickingtoad]

Ok.

I need to solve this. I got the equation from above from R(x)-C(x)=0 and need to find the break even points.

\(\displaystyle {-0.2x^{2}+166x-3550=0\)

I need to use the quadratic formula. I'm not getting this right though... What am I doing wrong?

Quadratic formula
$\displaystyle \frac{x=-b\pm\sqrt{b^{2}-4ac}}{2a}$

1. $\displaystyle \frac{-166\pm\sqrt{166^{2}-4(-0.2)(-3550)}}{2(-0.2)}$

2. $\displaystyle \frac{-166\pm\sqrt{27556-2840}}{-0.4}$

3. $\displaystyle \frac{-166\pm\sqrt{24716}}{-0.4}$

This comes out as incorrect. What is wrong?

 

[galactus]

You have both solutions. One is 21.97

The other is 808.03.

You do know that in the quadratic formula there are two solutions?. Hence, the plus and minus cases.

 

[kickingtoad]

It was correct when I used decimals. Thx.