32/y=3+y-4

Solution for 32/y=3+y-4 equation:

32/y=3+y-4

We move all terms to the left:

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


Domain of the equation: y!=0

y∈R


We add all the numbers together, and all the variables

32/y-(y-1)=0

We get rid of parentheses

32/y-y+1=0

We multiply all the terms by the denominator


-y*y+1*y+32=0

We add all the numbers together, and all the variables

y-y*y+32=0

Wy multiply elements

-1y^2+y+32=0

a = -1; b = 1; c = +32;
Δ = b2-4ac
Δ = 12-4·(-1)·32
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{129}}{2*-1}=\frac{-1-\sqrt{129}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*-1}=\frac{-1+\sqrt{129}}{-2}
$