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

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

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

We move all terms to the left:

1r-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

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

r*2r-3*2r=0

Wy multiply elements

2r^2-6r=0

a = 2; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·2·0
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*2}=\frac{0}{4}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*2}=\frac{12}{4}
=3
$