1/5x+3=2x+42

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

1/5x+3=2x+42

We move all terms to the left:

1/5x+3-(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+3=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-10x^2-195x+1=0

a = -10; b = -195; c = +1;
Δ = b2-4ac
Δ = -1952-4·(-10)·1
Δ = 38065
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-195)-\sqrt{38065}}{2*-10}=\frac{195-\sqrt{38065}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-195)+\sqrt{38065}}{2*-10}=\frac{195+\sqrt{38065}}{-20}
$