4+(1/3r)=r-8

Solution for 4+(1/3r)=r-8 equation:

4+(1/3r)=r-8

We move all terms to the left:

4+(1/3r)-(r-8)=0


Domain of the equation: 3r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(+1/3r)-(r-8)+4=0

We get rid of parentheses

1/3r-r+8+4=0

We multiply all the terms by the denominator


-r*3r+8*3r+4*3r+1=0

Wy multiply elements

-3r^2+24r+12r+1=0

We add all the numbers together, and all the variables

-3r^2+36r+1=0

a = -3; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·(-3)·1
Δ = 1308
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:

$\sqrt{\Delta}=\sqrt{1308}=\sqrt{4*327}=\sqrt{4}*\sqrt{327}=2\sqrt{327}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{327}}{2*-3}=\frac{-36-2\sqrt{327}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{327}}{2*-3}=\frac{-36+2\sqrt{327}}{-6}
$