# Math Help - Entering Quadratic

$\frac{90}{s+5} + \frac{90}{s-5} = 2\frac{1}{8}$

If I want to enter this into quadratic formula, do I need to change it someway? The coefficients are in fractions and that's what confuses me.

2. Hi reallylongnicknam.
put the equation into this form a^2 +bx +C = 0

bjh

3. Firstly, multiple both sides by (s+5)(s-5). What do you get?

4. $s^2 -25$ I'm thinking there is cross multipling with those 90s.

5. I mean $\displaystyle \frac{90}{s+5} \times (s+5)(s-5)+ \frac{90}{s-5}\times (s+5)(s-5) = 2\frac{1}{8}\times (s+5)(s-5)$

What do you get?

6. $90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5)$

7. Good, now expand each term and group where they are like.

What do you get?

8. Originally Posted by reallylongnickname
$90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5)$
OK so far...continue...easier if you change 2 1/8 to 17/8

9. $-\frac{17}{8 }s^2 + 180s + 25 = 0$

Would I divide all coefficients by 8 to remove the fraction before I can plug this into the quadratic formula?

10. You don't have to: a, b, and c in " $ax^2+ bx+ c= 0$" don't have to be integers but you will probably find the arithmetic easier.

11. Originally Posted by reallylongnickname
$-\frac{17}{8 }s^2 + 180s + 25 = 0$
That is not correct.
90(S - 5) + 90(S + 5) = 180S

17/8(S - 5)(S + 5) = 17/8(S^2 - 25)

So: (17/8)S^2 - (17/8)25 - 180S = 0
17S^2 / 8 - 180S - 425 / 8 = 0
Multiply by 8:
17S^2 - 1440S - 425 = 0

OK?