I'm quite clueless as to how to solve this. I know the quadratic equation but am not sure how to get this equation into the format to plug in to the quadratic...

Solve the quadratic equation using the formula [1/(1+x)]-[1/(3-x)]=(6/35)
Show your work

multiply both sides by 35(x+1)(x-3)

then
35(x-3)+35(x+1)=6(x-3)(x+1)

and you should be able to assemble it from there.

so should i just multiply that all out and then plug it in?

Write the two terms on the left with a common denomimnator. Then convert to standard quadratic form.

[(3-x) - (1+x)]/[(1+x)(3-x)] = 6/35
(2-2x)/[(1+x)(3-x)] = 6/35
(1-x) = (3/35)[(1+x)(3-x)]
35 -35x = 3[-x^2 +2x +3)
3x^2 -41x +26 = 0
(3x -2)(x-13)
Check my work. No guarantees

We were lucky that one factored fairly easily. x = 2/3 or 13

i got the same answer but used the quadratic :) thank you for all your help everyone!

To solve the quadratic equation, we first need to rewrite it in the form ax^2 + bx + c = 0. In this case, we have the equation:

[1/(1+x)] - [1/(3-x)] = (6/35)

Let's simplify the equation step by step to bring it into the required form:

1/(1+x) - 1/(3-x) = 6/35

Multiply every term by the least common denominator (LCD), which is (1+x)(3-x):

[(3-x) - (1+x)](1+x)(3-x) = (6/35)(1+x)(3-x)

Simplifying further, we get:

(3-x - 1-x)(1+x)(3-x) = (6/35)(1+x)(3-x)

(-2)(1+x)(3-x) = (6/35)(1+x)(3-x)

Divide both sides of the equation by (1+x)(3-x):

-2 = (6/35)

We can see that the equation -2 = (6/35) is not true. Therefore, there is no solution to the given equation.