1/y+2=y-3/y

Solution for 1/y+2=y-3/y equation:

1/y+2=y-3/y

We move all terms to the left:

1/y+2-(y-3/y)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/y-(+y-3/y)+2=0

We get rid of parentheses

1/y-y+3/y+2=0

We multiply all the terms by the denominator


-y*y+2*y+1+3=0

We add all the numbers together, and all the variables

2y-y*y+4=0

Wy multiply elements

-1y^2+2y+4=0

a = -1; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-1)·4
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-1}=\frac{-2-2\sqrt{5}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-1}=\frac{-2+2\sqrt{5}}{-2}
$