(10/(2x-8))+(x/(x^2-4))=0

Solution for (10/(2x-8))+(x/(x^2-4))=0 equation:

(10/(2x-8))+(x/(x^2-4))=0


Domain of the equation: (2x-8))!=0

x∈R



Domain of the equation: (x^2-4))!=0

x∈R


We calculate fractions


(10x^2-40)/((2x-8))*x^2+(2x^2-8x)/((2x-8))*x^2=0

We multiply all the terms by the denominator


(10x^2-40)+(2x^2-8x)=0

We get rid of parentheses

10x^2+2x^2-8x-40=0

We add all the numbers together, and all the variables

12x^2-8x-40=0

a = 12; b = -8; c = -40;
Δ = b2-4ac
Δ = -82-4·12·(-40)
Δ = 1984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{31}}{2*12}=\frac{8-8\sqrt{31}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{31}}{2*12}=\frac{8+8\sqrt{31}}{24}
$