1/6r-(-1/3r)=-3

Solution for 1/6r-(-1/3r)=-3 equation:

1/6r-(-1/3r)=-3

We move all terms to the left:

1/6r-(-1/3r)-(-3)=0


Domain of the equation: 6r!=0

r!=0/6

r!=0

r∈R



Domain of the equation: 3r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

1/6r-(-1/3r)+3=0

We get rid of parentheses

1/6r+1/3r+3=0

We calculate fractions


3r/18r^2+6r/18r^2+3=0

We multiply all the terms by the denominator


3r+6r+3*18r^2=0

We add all the numbers together, and all the variables

9r+3*18r^2=0

Wy multiply elements

54r^2+9r=0

a = 54; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·54·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*54}=\frac{-18}{108}
=-1/6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*54}=\frac{0}{108}
=0
$