quadratic equation ?

do you mean to have brackets?

3/(x - 1) + 5/(x + 2) = 1
3(x + 2) + 5(x - 1) = (x - 1)(x + 2)
3x + 6 + 5x - 5 = x² + x - 2
x² - 7x - 3 = 0

x = (7 ± √((-7)² - 4(-3)(1)) / 2
= (7 ± √(61)) / 2

I like it. You left out your brackets and so did Janson N

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multiply both sides by (x-1)(x+2)
3x+6+5x-5=x^2+x-2
x^2-7x-3=0
x=7+-sqrt(61)/2

3/x x - 1 x + 5/x x + 2 x = 1 x

3 - x + 5 + 2x = x

undefined....

AS WRITTEN, question reads as :-
(3/x) - 1 + (5/x) + 2 = 1
3 - x + 5 + 2x = x
x + 8 = x
No solution.
If you mean :-
3 / (x - 1) + 5 / (x + 2) = 1 then you should, with respect, say so!

3(x + 2) + 5(x - 1) = (x + 2)(x - 1)
3x + 6 + 5x - 5 = x² + x - 2
8x + 1 = x² + x - 2
x² - 7x - 3 = 0
x = [ 7 ± √ (49 + 12) ] / 2
x = [ 7 ± √ (61) ] / 2
x = [ 7 ± √ (61) ] / 2
x = 7.41 , x = -3.41

Please note that this is not a quadratic equation, because these are representable in the format: y=a(x**2)+bx+c, such that "x" may have either 1 or 2 legitimate solutions.

This formula has ZERO legitimate solutions. Mathematicians say that the answer is "indeterminate". Ordinary people simply say that there is no solution for "x" in this equation. This is a trick question in algebra. Shame on whoever asked you this one.

It simplifies into 8/x = O, thus no real, imaginary, or complex value for x is correct.

my friend , it is not a quadratric equation :
8/x = 0 that means that x → ∞

3/x - 1 + 5/x + 2 = 1
3/x + 5/x = 1 - 2 + 1
Combine like terms
8/x = 0
And you can't solve it :E Eight divided anything will not give 0.
Did you leave out something anywhere?

multiply both sides by (x-1)(x) to get rid of the denominator. then combine like terms and use the quadratic (which I'm guess you know what it is right?) (negative b plus or minus radical b squared minus four a c over two a)

3/x-1+5/x+2=1
3-x+5+2x=x
x+8=x
there where no solve

3/(x - 1) + 5/(x + 2) = 1
(x - 1)(x + 2)[3/(x - 1) + 5/(x + 2)] = 1(x - 1)(x + 2)
3(x + 2) + 5(x - 1) = (x - 1)(x + 2)
3*x + 3*2 + 5*x - 5*1 = x*x - 1*x + x*2 - 1*2
3x + 6 + 5x - 5 = x^2 - x + 2x - 2
x^2 + x - 3x - 5x - 2 - 6 + 5 = 0
x^2 - 7x - 3 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -7
c = -3

x = [7 ±√(49 + 12)]/2
x = [7 ±√61]/2
x = [7 ±7.81]/2 (approx.)

x = [7 + 7.81]/2
x = 14.81/2
x = 7.405

x = [7 - 7.81]/2
x = -0.81/2
x = -0.405

∴ x = -0.405 , 7.405 (approx.)