(3y+1)2y=21

Solution for (3y+1)2y=21 equation:

(3y+1)2y=21

We move all terms to the left:

(3y+1)2y-(21)=0

We multiply parentheses

6y^2+2y-21=0

a = 6; b = 2; c = -21;
Δ = b2-4ac
Δ = 22-4·6·(-21)
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{127}}{2*6}=\frac{-2-2\sqrt{127}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{127}}{2*6}=\frac{-2+2\sqrt{127}}{12}
$