3(r-2)=4(3+1/2r)

Solution for 3(r-2)=4(3+1/2r) equation:

3(r-2)=4(3+1/2r)

We move all terms to the left:

3(r-2)-(4(3+1/2r))=0


Domain of the equation: 2r))!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

3(r-2)-(4(1/2r+3))=0

We multiply parentheses

3r-(4(1/2r+3))-6=0

We multiply all the terms by the denominator


3r*2r-6*2r+3))-(4(1+3))=0

We add all the numbers together, and all the variables

3r*2r-6*2r+3))-(44)=0

We add all the numbers together, and all the variables

3r*2r-6*2r=0

Wy multiply elements

6r^2-12r=0

a = 6; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·6·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*6}=\frac{0}{12}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*6}=\frac{24}{12}
=2
$