2/3y+12=7/5y-20

Solution for 2/3y+12=7/5y-20 equation:

2/3y+12=7/5y-20

We move all terms to the left:

2/3y+12-(7/5y-20)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 5y-20)!=0

y∈R


We get rid of parentheses

2/3y-7/5y+20+12=0

We calculate fractions


10y/15y^2+(-21y)/15y^2+20+12=0

We add all the numbers together, and all the variables

10y/15y^2+(-21y)/15y^2+32=0

We multiply all the terms by the denominator


10y+(-21y)+32*15y^2=0

Wy multiply elements

480y^2+10y+(-21y)=0

We get rid of parentheses

480y^2+10y-21y=0

We add all the numbers together, and all the variables

480y^2-11y=0

a = 480; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·480·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*480}=\frac{0}{960}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*480}=\frac{22}{960}
=11/480
$