3(-2r)=30(-r-2)+r/6

Solution for 3(-2r)=30(-r-2)+r/6 equation:

3(-2r)=30(-r-2)+r/6

We move all terms to the left:

3(-2r)-(30(-r-2)+r/6)=0

We add all the numbers together, and all the variables

3(-2r)-(30(-1r-2)+r/6)=0

We multiply parentheses

-6r-(30(-1r-2)+r/6)=0

We multiply all the terms by the denominator


-6r*6)-(30(-1r-2)+r=0

We add all the numbers together, and all the variables

r-6r*6)-(30(-1r-2)=0

Wy multiply elements

-36r^2+r=0

a = -36; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-36)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-36}=\frac{-2}{-72}
=1/36
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-36}=\frac{0}{-72}
=0
$