(2/3r)-4=16+2r

Solution for (2/3r)-4=16+2r equation:

(2/3r)-4=16+2r

We move all terms to the left:

(2/3r)-4-(16+2r)=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)-(2r+16)-4=0

We get rid of parentheses

2/3r-2r-16-4=0

We multiply all the terms by the denominator


-2r*3r-16*3r-4*3r+2=0

Wy multiply elements

-6r^2-48r-12r+2=0

We add all the numbers together, and all the variables

-6r^2-60r+2=0

a = -6; b = -60; c = +2;
Δ = b2-4ac
Δ = -602-4·(-6)·2
Δ = 3648
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:

$\sqrt{\Delta}=\sqrt{3648}=\sqrt{64*57}=\sqrt{64}*\sqrt{57}=8\sqrt{57}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-8\sqrt{57}}{2*-6}=\frac{60-8\sqrt{57}}{-12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+8\sqrt{57}}{2*-6}=\frac{60+8\sqrt{57}}{-12}
$