3/x-6-5/7x-42=3

Solution for 3/x-6-5/7x-42=3 equation:

3/x-6-5/7x-42=3

We move all terms to the left:

3/x-6-5/7x-42-(3)=0


Domain of the equation: x!=0

x∈R



Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R


We add all the numbers together, and all the variables

3/x-5/7x-51=0

We calculate fractions


21x/7x^2+(-5x)/7x^2-51=0

We multiply all the terms by the denominator


21x+(-5x)-51*7x^2=0

Wy multiply elements

-357x^2+21x+(-5x)=0

We get rid of parentheses

-357x^2+21x-5x=0

We add all the numbers together, and all the variables

-357x^2+16x=0

a = -357; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-357)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-357}=\frac{-32}{-714}
=16/357
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-357}=\frac{0}{-714}
=0
$