(4/x-6)+(2/2x-12)=10

Solution for (4/x-6)+(2/2x-12)=10 equation:

(4/x-6)+(2/2x-12)=10

We move all terms to the left:

(4/x-6)+(2/2x-12)-(10)=0


Domain of the equation: x-6)!=0

x∈R



Domain of the equation: 2x-12)!=0

x∈R


We get rid of parentheses

4/x+2/2x-6-12-10=0

We calculate fractions


8x/2x^2+2x/2x^2-6-12-10=0

We add all the numbers together, and all the variables

8x/2x^2+2x/2x^2-28=0

We multiply all the terms by the denominator


8x+2x-28*2x^2=0

We add all the numbers together, and all the variables

10x-28*2x^2=0

Wy multiply elements

-56x^2+10x=0

a = -56; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-56)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-56}=\frac{-20}{-112}
=5/28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-56}=\frac{0}{-112}
=0
$