7y(2y^-3)=2

Solution for 7y(2y^-3)=2 equation:

7y(2y^-3)=2

We move all terms to the left:

7y(2y^-3)-(2)=0

We multiply parentheses

14y^2-21y-2=0

a = 14; b = -21; c = -2;
Δ = b2-4ac
Δ = -212-4·14·(-2)
Δ = 553
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{-(-21)-\sqrt{553}}{2*14}=\frac{21-\sqrt{553}}{28}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{553}}{2*14}=\frac{21+\sqrt{553}}{28}
$