1/3r-2=3r+5

Solution for 1/3r-2=3r+5 equation:

1/3r-2=3r+5

We move all terms to the left:

1/3r-2-(3r+5)=0


Domain of the equation: 3r!=0

r!=0/3

r!=0

r∈R


We get rid of parentheses

1/3r-3r-5-2=0

We multiply all the terms by the denominator


-3r*3r-5*3r-2*3r+1=0

Wy multiply elements

-9r^2-15r-6r+1=0

We add all the numbers together, and all the variables

-9r^2-21r+1=0

a = -9; b = -21; c = +1;
Δ = b2-4ac
Δ = -212-4·(-9)·1
Δ = 477
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:

$\sqrt{\Delta}=\sqrt{477}=\sqrt{9*53}=\sqrt{9}*\sqrt{53}=3\sqrt{53}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{53}}{2*-9}=\frac{21-3\sqrt{53}}{-18}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{53}}{2*-9}=\frac{21+3\sqrt{53}}{-18}
$