32-(3/8*y)=3/4y+5

Solution for 32-(3/8*y)=3/4y+5 equation:

32-(3/8y)=3/4y+5

We move all terms to the left:

32-(3/8y)-(3/4y+5)=0


Domain of the equation: 8y)!=0

y!=0/1

y!=0

y∈R



Domain of the equation: 4y+5)!=0

y∈R


We add all the numbers together, and all the variables

-(+3/8y)-(3/4y+5)+32=0

We get rid of parentheses

-3/8y-3/4y-5+32=0

We calculate fractions


(-12y)/32y^2+(-24y)/32y^2-5+32=0

We add all the numbers together, and all the variables

(-12y)/32y^2+(-24y)/32y^2+27=0

We multiply all the terms by the denominator


(-12y)+(-24y)+27*32y^2=0

Wy multiply elements

864y^2+(-12y)+(-24y)=0

We get rid of parentheses

864y^2-12y-24y=0

We add all the numbers together, and all the variables

864y^2-36y=0

a = 864; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·864·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*864}=\frac{0}{1728}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*864}=\frac{72}{1728}
=1/24
$