(3/5x)-20=(20*2)+x

Solution for (3/5x)-20=(20*2)+x equation:

(3/5x)-20=(20*2)+x

We move all terms to the left:

(3/5x)-20-((20*2)+x)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/5x)-(40+x)-20=0

We get rid of parentheses

3/5x-x-40-20=0

We multiply all the terms by the denominator


-x*5x-40*5x-20*5x+3=0

Wy multiply elements

-5x^2-200x-100x+3=0

We add all the numbers together, and all the variables

-5x^2-300x+3=0

a = -5; b = -300; c = +3;
Δ = b2-4ac
Δ = -3002-4·(-5)·3
Δ = 90060
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:

$\sqrt{\Delta}=\sqrt{90060}=\sqrt{4*22515}=\sqrt{4}*\sqrt{22515}=2\sqrt{22515}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-2\sqrt{22515}}{2*-5}=\frac{300-2\sqrt{22515}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+2\sqrt{22515}}{2*-5}=\frac{300+2\sqrt{22515}}{-10}
$