1/7x-1/3x+6=0

Solution for 1/7x-1/3x+6=0 equation:

1/7x-1/3x+6=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We calculate fractions


3x/21x^2+(-7x)/21x^2+6=0

We multiply all the terms by the denominator


3x+(-7x)+6*21x^2=0

Wy multiply elements

126x^2+3x+(-7x)=0

We get rid of parentheses

126x^2+3x-7x=0

We add all the numbers together, and all the variables

126x^2-4x=0

a = 126; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·126·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*126}=\frac{0}{252}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*126}=\frac{8}{252}
=2/63
$