If you mean 2/(x+4) + 4/(x-5) = 28/( (x+2)(x-5) ) then:
Multiply both sides by (x+4)(x+2)(x-5)
2(x+2)(x-5) + 4(x+4)(x+2) = 28(x+4)
x(x^2 - 3x - 10) + 4(x^2 + 6x + 8) = 28x + 112
x^3 - 3x^2 - 10x + 4x^2 + 24x + 32 = 28x + 112
x^3 + x^2 + 14x + 32 = 28x + 112
x^3 + x^2 - 14x - 80 = 0
Checking for nice solutions. . .
80: 2 * 2 * 2 * 2 * 5
What you can make with the prime factors of 80: 2, 4, 5, 8, 10, and a bunch of larger numbers.
If you plug 10 in for x, the x^3 makes 1000, much larger than the rest of the equation, so 10 is not a root of x^3 + x^2 - 14x - 80 = 0
2: -96
4: -56
5: 125 + 25 - 70 - 80 = 0
x = 5 is a root
(x - 5)( ? ) = x^3 + x^2 - 14x - 80
(x-5)(x^2 + ?x + 16) = x^3 + x^2 - 14x - 80
The -14x is made of 16x + (-5)(?x)
-14x = 16x -5?x
-14 = 16 - 5?
-30 = -5?
? = 6
(x-5)(x^2 + 6x + 16) = x^3 + x^2 - 14x - 80
Checking (x-5)(x^2 + 6x + 16). . .
x^3 + 6x^2 + 16x -5x^2 - 30x - 80
x^3 + x^2 - 14x - 80 good
(x-5)(x^2 + 6x + 16)
factor x^2 + 6x + 16
If you're in pre-algebra, you won't have run into it yet, but there's this thing called the quadratic formula for solving and factoring things with x^2
If you have ax^2 + bx + c = 0 then the values for x are x = (-b +- sqrt(b^2 - 4ac) ) / 2a
x^3 + x^2 - 14x - 80 = (x-5)(x^2 + 6x + 16) = 0
a = 1, b = 6, c = 16
x = (-6 +- sqrt(6^2 - 4(1)(16)) ) / 2(1)
x = (-6 +- sqrt(36 - 64) ) / 2
x = (-6 +- sqrt(-28)) / 2
You can't take the square root of a negative number, so doesn't factor.
This means there is no way to make x^2 + 6x + 16 = 0.
That means the only way to make (x-5)(x^2 + 6x + 16) = 0 is if x - 5 = 0, which gives us x = 5
Answer: x = 5