8(y+2)/2y=y=-2

Solution for 8(y+2)/2y=y=-2 equation:

8(y+2)/2y=y=-2

We move all terms to the left:

8(y+2)/2y-(y)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R


We add all the numbers together, and all the variables

-1y+8(y+2)/2y=0

We multiply all the terms by the denominator


-1y*2y+8(y+2)=0

We multiply parentheses

-1y*2y+8y+16=0

Wy multiply elements

-2y^2+8y+16=0

a = -2; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·(-2)·16
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{3}}{2*-2}=\frac{-8-8\sqrt{3}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{3}}{2*-2}=\frac{-8+8\sqrt{3}}{-4}
$