(13-2/3r)-(1-r)=

Solution for (13-2/3r)-(1-r)= equation:

(13-2/3r)-(1-r)=

We move all terms to the left:

(13-2/3r)-(1-r)-()=0


Domain of the equation: 3r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(-2/3r+13)-(-1r+1)-()=0

We add all the numbers together, and all the variables

(-2/3r+13)-(-1r+1)=0

We get rid of parentheses

-2/3r+1r+13-1=0

We multiply all the terms by the denominator


1r*3r+13*3r-1*3r-2=0

Wy multiply elements

3r^2+39r-3r-2=0

We add all the numbers together, and all the variables

3r^2+36r-2=0

a = 3; b = 36; c = -2;
Δ = b2-4ac
Δ = 362-4·3·(-2)
Δ = 1320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1320}=\sqrt{4*330}=\sqrt{4}*\sqrt{330}=2\sqrt{330}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{330}}{2*3}=\frac{-36-2\sqrt{330}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{330}}{2*3}=\frac{-36+2\sqrt{330}}{6}
$