3/2y+12=7/3y-20

Solution for 3/2y+12=7/3y-20 equation:

3/2y+12=7/3y-20

We move all terms to the left:

3/2y+12-(7/3y-20)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 3y-20)!=0

y∈R


We get rid of parentheses

3/2y-7/3y+20+12=0

We calculate fractions


9y/6y^2+(-14y)/6y^2+20+12=0

We add all the numbers together, and all the variables

9y/6y^2+(-14y)/6y^2+32=0

We multiply all the terms by the denominator


9y+(-14y)+32*6y^2=0

Wy multiply elements

192y^2+9y+(-14y)=0

We get rid of parentheses

192y^2+9y-14y=0

We add all the numbers together, and all the variables

192y^2-5y=0

a = 192; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·192·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*192}=\frac{0}{384}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*192}=\frac{10}{384}
=5/192
$