0.5r-3=3(4-3/2r)

Solution for 0.5r-3=3(4-3/2r) equation:

0.5r-3=3(4-3/2r)

We move all terms to the left:

0.5r-3-(3(4-3/2r))=0


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

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

0.5r-(3(-3/2r+4))-3=0

We multiply all the terms by the denominator


(0.5r)*2r-3*2r-3+4))-(3(+4))=0

We add all the numbers together, and all the variables

(+0.5r)*2r-3*2r-3+4))-(34)=0

We add all the numbers together, and all the variables

(+0.5r)*2r-3*2r=0

We multiply parentheses

0r^2-3*2r=0

Wy multiply elements

0r^2-6r=0

We add all the numbers together, and all the variables

r^2-6r=0

a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2}
=6
$