2(4y+1)3y=12

Solution for 2(4y+1)3y=12 equation:

2(4y+1)3y=12

We move all terms to the left:

2(4y+1)3y-(12)=0

We multiply parentheses

24y^2+6y-12=0

a = 24; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·24·(-12)
Δ = 1188
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:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{33}}{2*24}=\frac{-6-6\sqrt{33}}{48}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{33}}{2*24}=\frac{-6+6\sqrt{33}}{48}
$