1/5x=2x+42

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

1/5x=2x+42

We move all terms to the left:

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


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We get rid of parentheses

1/5x-2x-42=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

a = -10; b = -210; c = +1;
Δ = b2-4ac
Δ = -2102-4·(-10)·1
Δ = 44140
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:

$\sqrt{\Delta}=\sqrt{44140}=\sqrt{4*11035}=\sqrt{4}*\sqrt{11035}=2\sqrt{11035}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-2\sqrt{11035}}{2*-10}=\frac{210-2\sqrt{11035}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+2\sqrt{11035}}{2*-10}=\frac{210+2\sqrt{11035}}{-20}
$