(1/5x)+(3x)=(2x)+42

Solution for (1/5x)+(3x)=(2x)+42 equation:

(1/5x)+(3x)=(2x)+42

We move all terms to the left:

(1/5x)+(3x)-((2x)+42)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/5x)+3x-(2x+42)=0

We add all the numbers together, and all the variables

3x+(+1/5x)-(2x+42)=0

We get rid of parentheses

3x+1/5x-2x-42=0

We multiply all the terms by the denominator


3x*5x-2x*5x-42*5x+1=0

Wy multiply elements

15x^2-10x^2-210x+1=0

We add all the numbers together, and all the variables

5x^2-210x+1=0

a = 5; b = -210; c = +1;
Δ = b2-4ac
Δ = -2102-4·5·1
Δ = 44080
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:

$\sqrt{\Delta}=\sqrt{44080}=\sqrt{16*2755}=\sqrt{16}*\sqrt{2755}=4\sqrt{2755}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-4\sqrt{2755}}{2*5}=\frac{210-4\sqrt{2755}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+4\sqrt{2755}}{2*5}=\frac{210+4\sqrt{2755}}{10}
$