1/2r-3/2+2=1/3r-3

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

1/2r-3/2+2=1/3r-3

We move all terms to the left:

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


Domain of the equation: 2r!=0

r!=0/2

r!=0

r∈R



Domain of the equation: 3r-3)!=0

r∈R


We get rid of parentheses

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

We calculate fractions


3r/24r^2+(-8r)/24r^2+(-9r)/24r^2+3+2=0

We add all the numbers together, and all the variables

3r/24r^2+(-8r)/24r^2+(-9r)/24r^2+5=0

We multiply all the terms by the denominator


3r+(-8r)+(-9r)+5*24r^2=0

Wy multiply elements

120r^2+3r+(-8r)+(-9r)=0

We get rid of parentheses

120r^2+3r-8r-9r=0

We add all the numbers together, and all the variables

120r^2-14r=0

a = 120; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·120·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*120}=\frac{0}{240}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*120}=\frac{28}{240}
=7/60
$