-(5/3x)+6/2x+1=0

Solution for -(5/3x)+6/2x+1=0 equation:

-(5/3x)+6/2x+1=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

-(+5/3x)+6/2x+1=0

We get rid of parentheses

-5/3x+6/2x+1=0

We calculate fractions


(-10x)/6x^2+18x/6x^2+1=0

We multiply all the terms by the denominator


(-10x)+18x+1*6x^2=0

We add all the numbers together, and all the variables

18x+(-10x)+1*6x^2=0

Wy multiply elements

6x^2+18x+(-10x)=0

We get rid of parentheses

6x^2+18x-10x=0

We add all the numbers together, and all the variables

6x^2+8x=0

a = 6; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·6·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*6}=\frac{0}{12}
=0
$