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

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

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

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



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

y∈R


We get rid of parentheses

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

We calculate fractions


15y/10y^2+(-14y)/10y^2+20+12=0

We add all the numbers together, and all the variables

15y/10y^2+(-14y)/10y^2+32=0

We multiply all the terms by the denominator


15y+(-14y)+32*10y^2=0

Wy multiply elements

320y^2+15y+(-14y)=0

We get rid of parentheses

320y^2+15y-14y=0

We add all the numbers together, and all the variables

320y^2+y=0

a = 320; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·320·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*320}=\frac{-2}{640}
=-1/320
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*320}=\frac{0}{640}
=0
$