(6+x)+108=12/5x

Solution for (6+x)+108=12/5x equation:

(6+x)+108=12/5x

We move all terms to the left:

(6+x)+108-(12/5x)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(x+6)-(+12/5x)+108=0

We get rid of parentheses

x-12/5x+6+108=0

We multiply all the terms by the denominator


x*5x+6*5x+108*5x-12=0

Wy multiply elements

5x^2+30x+540x-12=0

We add all the numbers together, and all the variables

5x^2+570x-12=0

a = 5; b = 570; c = -12;
Δ = b2-4ac
Δ = 5702-4·5·(-12)
Δ = 325140
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:

$\sqrt{\Delta}=\sqrt{325140}=\sqrt{4*81285}=\sqrt{4}*\sqrt{81285}=2\sqrt{81285}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(570)-2\sqrt{81285}}{2*5}=\frac{-570-2\sqrt{81285}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(570)+2\sqrt{81285}}{2*5}=\frac{-570+2\sqrt{81285}}{10}
$