2(y^2+1/y^2)-9(y+1/y)=8

Solution for 2(y^2+1/y^2)-9(y+1/y)=8 equation:

2(y^2+1/y^2)-9(y+1/y)=8

We move all terms to the left:

2(y^2+1/y^2)-9(y+1/y)-(8)=0


Domain of the equation: y^2)!=0

y!=0/1

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

2(y^2+1/y^2)-9(+y+1/y)-8=0

We multiply parentheses

2y^2+2y-9y-9y-8=0

We add all the numbers together, and all the variables

2y^2-16y-8=0

a = 2; b = -16; c = -8;
Δ = b2-4ac
Δ = -162-4·2·(-8)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{5}}{2*2}=\frac{16-8\sqrt{5}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{5}}{2*2}=\frac{16+8\sqrt{5}}{4}
$