3/2y+12=71/5y-20

Solution for 3/2y+12=71/5y-20 equation:

3/2y+12=71/5y-20

We move all terms to the left:

3/2y+12-(71/5y-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-71/5y+20+12=0

We calculate fractions


15y/10y^2+(-142y)/10y^2+20+12=0

We add all the numbers together, and all the variables

15y/10y^2+(-142y)/10y^2+32=0

We multiply all the terms by the denominator


15y+(-142y)+32*10y^2=0

Wy multiply elements

320y^2+15y+(-142y)=0

We get rid of parentheses

320y^2+15y-142y=0

We add all the numbers together, and all the variables

320y^2-127y=0

a = 320; b = -127; c = 0;
Δ = b2-4ac
Δ = -1272-4·320·0
Δ = 16129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16129}=127$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-127)-127}{2*320}=\frac{0}{640}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+127}{2*320}=\frac{254}{640}
=127/320
$