(6+y)/y-y/2=3

Solution for (6+y)/y-y/2=3 equation:

(6+y)/y-y/2=3

We move all terms to the left:

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


Domain of the equation: y!=0

y∈R


We add all the numbers together, and all the variables

(y+6)/y-y/2-3=0

We calculate fractions


(-1y^2)/2y+(2y+12)/2y-3=0

We multiply all the terms by the denominator


(-1y^2)+(2y+12)-3*2y=0

Wy multiply elements

(-1y^2)+(2y+12)-6y=0

We get rid of parentheses

-1y^2+2y-6y+12=0

We add all the numbers together, and all the variables

-1y^2-4y+12=0

a = -1; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-1}=\frac{-4}{-2}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-1}=\frac{12}{-2}
=-6
$